Limit as $x$ tend to zero of: $x/[ln (x^2+2x+4) - ln(x+4)]$
Without making use of LHôpital's Rule solve:
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.
limits logarithms rational-functions
asked Mar 30 '15 at 23:55
Paulo Martins
136
add a comment |
Without making use of LHôpital's Rule solve:
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.
limits logarithms rational-functions
asked Mar 30 '15 at 23:55
Paulo Martins
136
add a comment |
Without making use of LHôpital's Rule solve:
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.
limits logarithms rational-functions
asked Mar 30 '15 at 23:55
Paulo Martins
136
Without making use of LHôpital's Rule solve:
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.
limits logarithms rational-functions
limits logarithms rational-functions
asked Mar 30 '15 at 23:55
Paulo Martins
136
asked Mar 30 '15 at 23:55
Paulo Martins
136
edited Mar 31 '15 at 0:37
asked Mar 30 '15 at 23:55
Paulo Martins
136
asked Mar 30 '15 at 23:55
Paulo Martins
136
asked Mar 30 '15 at 23:55
Paulo Martins
136
136
add a comment |
add a comment |
5 Answers
5
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oldest
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An approach without L'Hopital's rule.
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}=lim_{xto 0} {1over {1over x}ln {x^2+2x+4over x+4}}=lim_{xto 0} {1over ln big ({x^2+2x+4over x+4}big)^{1over x}}$$
but
$$({x^2+2x+4over x+4}big)^{1over x}=({x^2+x+x+4over x+4}big)^{1over x}=big(1+{x^2+xover x+4}big)^{1over x}=big(1+{ x(x+1)over x+4}big)^{1over x}=big(1+colorred x({x+1over x+4})big)^{1over colorred x}$$ which tends to $e^{1over 4}$ as $x$ goes to $0$. So the original limit is ${1over ln{e^{1over 4}}}=colorred 4$
answered Mar 31 '15 at 0:30
Fermat
4,3631926
I understand the process which you describe in it's broader form but I fail to grasp the transformation of the rational expression you describe in the second paragraph can you write it down?
– Paulo Martins
Mar 31 '15 at 0:44
I edited the answer and added some details. Is it clear now? @PauloMartins
– Fermat
Mar 31 '15 at 0:53
Because lim as n-> 0 of(1+1/n)^n = e correct? @Fermat
– Paulo Martins
Mar 31 '15 at 0:58
Yes but $$lim_{ntoinfty}(1+{1over n})^n=e$$ and in general $$lim_{ntoinfty}(1+{kover n})^n=e^k$$
– Fermat
Mar 31 '15 at 0:59
add a comment |
By L'Hospital we have
$$lim_{xto 0} frac{1}{frac{2x + 2}{x^2 + 2x + 4} - frac{1}{x+4}} = color{red} 4$$
answered Mar 31 '15 at 0:14
Aaron Maroja
15.5k51445
add a comment |
Outline: Flip it over, and calculate
$$lim_{xto 0} frac{ln(x^2+2x+4)-ln(4)}{x} -lim_{xto 0}frac{ln(x+4)-ln(4)}{x}.$$
We recognize the two limits as derivatives at $0$ of $ln(x^2+2x+4)$ and $ln(x+4)$ respectively.
answered Mar 31 '15 at 0:34
André Nicolas
451k36422806
Yes, thank you. For the sake of familiarity I had used $h$, then decided not to, but did not manage to change all of them.
– André Nicolas
Mar 31 '15 at 0:41
Thanks for your reply - in any case, nice answer!
– user84413
Mar 31 '15 at 0:42
You are welcome. I consider the Taylor series approach the natural one, because it directly addresses the behaviour of the functions near $0$. But because of the order in which such things are done in standard North American calculus courses, Taylor series might be dealt with after this kind of limit problem.
– André Nicolas
Mar 31 '15 at 0:46
What is the property that makes this particular and other limits in general "fliparoveble"
– Paulo Martins
Mar 31 '15 at 1:16
They basically all are. We have $lim{xto a}f(x)=Lne 0$ if and only if $lim_{xto a}frac{1}{f(x)}=frac{1}{L}$. And we can even say useful things when $L= 0$, but $Lne 0$ in this problem.
– André Nicolas
Mar 31 '15 at 2:16
|
show 1 more comment
In the eternal words of Claude Leibovici, love Taylor Series.
We have
begin{align*}
ln(x^{2}+2x+4) &= ln(4) + frac{x}{2} + frac{x^{2}}{8} + mathcal{O}(x^{3})\
ln(x+4) &=ln(4) + frac{x}{4} -frac{x^{2}}{32} + mathcal{O}(x^{3}).
end{align*}
So
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)} = lim_{xto 0} frac{x}{ln(4) + frac{x}{2} + frac{x^{2}}{8} - left( ln(4) + frac{x}{4} -frac{x^{2}}{32}right) + mathcal{O}(x^{3})} = lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})}.$$
Divide the numerator and denominator by the lowest power of $x$ in the denominator to get
$$lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})} = lim_{xto 0} frac{frac{1}{x} x}{frac{1}{x}left( frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})right)} = lim_{xto 0}frac{1}{frac{1}{4} + frac{5x}{32} + mathcal{O}(x^{2})} = frac{1}{frac{1}{4}} = 4.$$
answered Mar 31 '15 at 0:38
JessicaK
4,46151733
In the final step of the Taylor series are you stating the inverse of the property lim x-> 00 of (a0x^n + a1x^n-1 ... an) = lim x->00 a0x^n? @JessicaK
– Paulo Martins
Mar 31 '15 at 1:00
@PauloMartins I am having some trouble understanding the equation that you wrote. I simply dropped the higher order terms of the polynomials because they do not affect the limit, but I edited my answer to avoid doing this.
– JessicaK
Mar 31 '15 at 1:03
@PauloMartins If "00" is supposed to be $infty$, hopefully by looking at the above modification, you should be able to convince yourself of an analogous property for limits of polynomials towards $0$ instead of $infty$.
– JessicaK
Mar 31 '15 at 1:13
Yes it was supposed to mean exactly that. And I have convinced myself of such property. Thanks
– Paulo Martins
Mar 31 '15 at 1:15
Thanks you, JessicaK for having noticed my love for Taylor ! Cheers :-)
– Claude Leibovici
Mar 31 '15 at 6:43
add a comment |
Rewrite it as
$$limlimits_{xrightarrow0}frac{x}{ln left( frac{x^2+2x+4}{x+4}right)}$$
Apply L'Hopital's rule:
$$limlimits_{xrightarrow0}frac{1}{frac{x^2+8x+4}{(x+4)(x^2+2x+4)}}$$
Then simply evaluate the limit:
$$frac{1}{frac{0^2+8times0+4}{(0+4)(0^2+2times0+4)}} = frac{1}{frac{4}{16}} = 4$$
answered Mar 31 '15 at 0:14
TokenToucan
2,4171419
Why not apply L'Hopital's rule with the difference of two logarithms rather than making it into a logarithm of a quotient first. That way you don't need the quotient rule. ${}qquad{}$
– Michael Hardy
Mar 31 '15 at 1:32
add a comment |
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5 Answers
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5 Answers
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An approach without L'Hopital's rule.
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}=lim_{xto 0} {1over {1over x}ln {x^2+2x+4over x+4}}=lim_{xto 0} {1over ln big ({x^2+2x+4over x+4}big)^{1over x}}$$
but
$$({x^2+2x+4over x+4}big)^{1over x}=({x^2+x+x+4over x+4}big)^{1over x}=big(1+{x^2+xover x+4}big)^{1over x}=big(1+{ x(x+1)over x+4}big)^{1over x}=big(1+colorred x({x+1over x+4})big)^{1over colorred x}$$ which tends to $e^{1over 4}$ as $x$ goes to $0$. So the original limit is ${1over ln{e^{1over 4}}}=colorred 4$
answered Mar 31 '15 at 0:30
Fermat
4,3631926
I understand the process which you describe in it's broader form but I fail to grasp the transformation of the rational expression you describe in the second paragraph can you write it down?
– Paulo Martins
Mar 31 '15 at 0:44
I edited the answer and added some details. Is it clear now? @PauloMartins
– Fermat
Mar 31 '15 at 0:53
Because lim as n-> 0 of(1+1/n)^n = e correct? @Fermat
– Paulo Martins
Mar 31 '15 at 0:58
Yes but $$lim_{ntoinfty}(1+{1over n})^n=e$$ and in general $$lim_{ntoinfty}(1+{kover n})^n=e^k$$
– Fermat
Mar 31 '15 at 0:59
add a comment |
An approach without L'Hopital's rule.
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}=lim_{xto 0} {1over {1over x}ln {x^2+2x+4over x+4}}=lim_{xto 0} {1over ln big ({x^2+2x+4over x+4}big)^{1over x}}$$
but
$$({x^2+2x+4over x+4}big)^{1over x}=({x^2+x+x+4over x+4}big)^{1over x}=big(1+{x^2+xover x+4}big)^{1over x}=big(1+{ x(x+1)over x+4}big)^{1over x}=big(1+colorred x({x+1over x+4})big)^{1over colorred x}$$ which tends to $e^{1over 4}$ as $x$ goes to $0$. So the original limit is ${1over ln{e^{1over 4}}}=colorred 4$
answered Mar 31 '15 at 0:30
Fermat
4,3631926
I understand the process which you describe in it's broader form but I fail to grasp the transformation of the rational expression you describe in the second paragraph can you write it down?
– Paulo Martins
Mar 31 '15 at 0:44
I edited the answer and added some details. Is it clear now? @PauloMartins
– Fermat
Mar 31 '15 at 0:53
Because lim as n-> 0 of(1+1/n)^n = e correct? @Fermat
– Paulo Martins
Mar 31 '15 at 0:58
Yes but $$lim_{ntoinfty}(1+{1over n})^n=e$$ and in general $$lim_{ntoinfty}(1+{kover n})^n=e^k$$
– Fermat
Mar 31 '15 at 0:59
add a comment |
An approach without L'Hopital's rule.
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}=lim_{xto 0} {1over {1over x}ln {x^2+2x+4over x+4}}=lim_{xto 0} {1over ln big ({x^2+2x+4over x+4}big)^{1over x}}$$
but
$$({x^2+2x+4over x+4}big)^{1over x}=({x^2+x+x+4over x+4}big)^{1over x}=big(1+{x^2+xover x+4}big)^{1over x}=big(1+{ x(x+1)over x+4}big)^{1over x}=big(1+colorred x({x+1over x+4})big)^{1over colorred x}$$ which tends to $e^{1over 4}$ as $x$ goes to $0$. So the original limit is ${1over ln{e^{1over 4}}}=colorred 4$
answered Mar 31 '15 at 0:30
Fermat
4,3631926
An approach without L'Hopital's rule.
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)}=lim_{xto 0} {1over {1over x}ln {x^2+2x+4over x+4}}=lim_{xto 0} {1over ln big ({x^2+2x+4over x+4}big)^{1over x}}$$
but
$$({x^2+2x+4over x+4}big)^{1over x}=({x^2+x+x+4over x+4}big)^{1over x}=big(1+{x^2+xover x+4}big)^{1over x}=big(1+{ x(x+1)over x+4}big)^{1over x}=big(1+colorred x({x+1over x+4})big)^{1over colorred x}$$ which tends to $e^{1over 4}$ as $x$ goes to $0$. So the original limit is ${1over ln{e^{1over 4}}}=colorred 4$
answered Mar 31 '15 at 0:30
Fermat
4,3631926
edited Nov 26 '18 at 17:37
answered Mar 31 '15 at 0:30
Fermat
4,3631926
answered Mar 31 '15 at 0:30
Fermat
4,3631926
answered Mar 31 '15 at 0:30
Fermat
4,3631926
4,3631926
I understand the process which you describe in it's broader form but I fail to grasp the transformation of the rational expression you describe in the second paragraph can you write it down?
– Paulo Martins
Mar 31 '15 at 0:44
I edited the answer and added some details. Is it clear now? @PauloMartins
– Fermat
Mar 31 '15 at 0:53
Because lim as n-> 0 of(1+1/n)^n = e correct? @Fermat
– Paulo Martins
Mar 31 '15 at 0:58
Yes but $$lim_{ntoinfty}(1+{1over n})^n=e$$ and in general $$lim_{ntoinfty}(1+{kover n})^n=e^k$$
– Fermat
Mar 31 '15 at 0:59
add a comment |
I understand the process which you describe in it's broader form but I fail to grasp the transformation of the rational expression you describe in the second paragraph can you write it down?
– Paulo Martins
Mar 31 '15 at 0:44
I edited the answer and added some details. Is it clear now? @PauloMartins
– Fermat
Mar 31 '15 at 0:53
Because lim as n-> 0 of(1+1/n)^n = e correct? @Fermat
– Paulo Martins
Mar 31 '15 at 0:58
Yes but $$lim_{ntoinfty}(1+{1over n})^n=e$$ and in general $$lim_{ntoinfty}(1+{kover n})^n=e^k$$
– Fermat
Mar 31 '15 at 0:59
I understand the process which you describe in it's broader form but I fail to grasp the transformation of the rational expression you describe in the second paragraph can you write it down?
– Paulo Martins
Mar 31 '15 at 0:44
I understand the process which you describe in it's broader form but I fail to grasp the transformation of the rational expression you describe in the second paragraph can you write it down?
– Paulo Martins
Mar 31 '15 at 0:44
I edited the answer and added some details. Is it clear now? @PauloMartins
– Fermat
Mar 31 '15 at 0:53
I edited the answer and added some details. Is it clear now? @PauloMartins
– Fermat
Mar 31 '15 at 0:53
Because lim as n-> 0 of(1+1/n)^n = e correct? @Fermat
– Paulo Martins
Mar 31 '15 at 0:58
Because lim as n-> 0 of(1+1/n)^n = e correct? @Fermat
– Paulo Martins
Mar 31 '15 at 0:58
Yes but $$lim_{ntoinfty}(1+{1over n})^n=e$$ and in general $$lim_{ntoinfty}(1+{kover n})^n=e^k$$
– Fermat
Mar 31 '15 at 0:59
Yes but $$lim_{ntoinfty}(1+{1over n})^n=e$$ and in general $$lim_{ntoinfty}(1+{kover n})^n=e^k$$
– Fermat
Mar 31 '15 at 0:59
add a comment |
By L'Hospital we have
$$lim_{xto 0} frac{1}{frac{2x + 2}{x^2 + 2x + 4} - frac{1}{x+4}} = color{red} 4$$
answered Mar 31 '15 at 0:14
Aaron Maroja
15.5k51445
add a comment |
By L'Hospital we have
$$lim_{xto 0} frac{1}{frac{2x + 2}{x^2 + 2x + 4} - frac{1}{x+4}} = color{red} 4$$
answered Mar 31 '15 at 0:14
Aaron Maroja
15.5k51445
add a comment |
By L'Hospital we have
$$lim_{xto 0} frac{1}{frac{2x + 2}{x^2 + 2x + 4} - frac{1}{x+4}} = color{red} 4$$
answered Mar 31 '15 at 0:14
Aaron Maroja
15.5k51445
By L'Hospital we have
$$lim_{xto 0} frac{1}{frac{2x + 2}{x^2 + 2x + 4} - frac{1}{x+4}} = color{red} 4$$
answered Mar 31 '15 at 0:14
Aaron Maroja
15.5k51445
answered Mar 31 '15 at 0:14
Aaron Maroja
15.5k51445
answered Mar 31 '15 at 0:14
Aaron Maroja
15.5k51445
answered Mar 31 '15 at 0:14
Aaron Maroja
15.5k51445
15.5k51445
add a comment |
add a comment |
Outline: Flip it over, and calculate
$$lim_{xto 0} frac{ln(x^2+2x+4)-ln(4)}{x} -lim_{xto 0}frac{ln(x+4)-ln(4)}{x}.$$
We recognize the two limits as derivatives at $0$ of $ln(x^2+2x+4)$ and $ln(x+4)$ respectively.
answered Mar 31 '15 at 0:34
André Nicolas
451k36422806
Yes, thank you. For the sake of familiarity I had used $h$, then decided not to, but did not manage to change all of them.
– André Nicolas
Mar 31 '15 at 0:41
Thanks for your reply - in any case, nice answer!
– user84413
Mar 31 '15 at 0:42
You are welcome. I consider the Taylor series approach the natural one, because it directly addresses the behaviour of the functions near $0$. But because of the order in which such things are done in standard North American calculus courses, Taylor series might be dealt with after this kind of limit problem.
– André Nicolas
Mar 31 '15 at 0:46
What is the property that makes this particular and other limits in general "fliparoveble"
– Paulo Martins
Mar 31 '15 at 1:16
They basically all are. We have $lim{xto a}f(x)=Lne 0$ if and only if $lim_{xto a}frac{1}{f(x)}=frac{1}{L}$. And we can even say useful things when $L= 0$, but $Lne 0$ in this problem.
– André Nicolas
Mar 31 '15 at 2:16
|
show 1 more comment
Outline: Flip it over, and calculate
$$lim_{xto 0} frac{ln(x^2+2x+4)-ln(4)}{x} -lim_{xto 0}frac{ln(x+4)-ln(4)}{x}.$$
We recognize the two limits as derivatives at $0$ of $ln(x^2+2x+4)$ and $ln(x+4)$ respectively.
answered Mar 31 '15 at 0:34
André Nicolas
451k36422806
Yes, thank you. For the sake of familiarity I had used $h$, then decided not to, but did not manage to change all of them.
– André Nicolas
Mar 31 '15 at 0:41
Thanks for your reply - in any case, nice answer!
– user84413
Mar 31 '15 at 0:42
You are welcome. I consider the Taylor series approach the natural one, because it directly addresses the behaviour of the functions near $0$. But because of the order in which such things are done in standard North American calculus courses, Taylor series might be dealt with after this kind of limit problem.
– André Nicolas
Mar 31 '15 at 0:46
What is the property that makes this particular and other limits in general "fliparoveble"
– Paulo Martins
Mar 31 '15 at 1:16
They basically all are. We have $lim{xto a}f(x)=Lne 0$ if and only if $lim_{xto a}frac{1}{f(x)}=frac{1}{L}$. And we can even say useful things when $L= 0$, but $Lne 0$ in this problem.
– André Nicolas
Mar 31 '15 at 2:16
|
show 1 more comment
Outline: Flip it over, and calculate
$$lim_{xto 0} frac{ln(x^2+2x+4)-ln(4)}{x} -lim_{xto 0}frac{ln(x+4)-ln(4)}{x}.$$
We recognize the two limits as derivatives at $0$ of $ln(x^2+2x+4)$ and $ln(x+4)$ respectively.
answered Mar 31 '15 at 0:34
André Nicolas
451k36422806
Outline: Flip it over, and calculate
$$lim_{xto 0} frac{ln(x^2+2x+4)-ln(4)}{x} -lim_{xto 0}frac{ln(x+4)-ln(4)}{x}.$$
We recognize the two limits as derivatives at $0$ of $ln(x^2+2x+4)$ and $ln(x+4)$ respectively.
answered Mar 31 '15 at 0:34
André Nicolas
451k36422806
edited Mar 31 '15 at 0:40
answered Mar 31 '15 at 0:34
André Nicolas
451k36422806
answered Mar 31 '15 at 0:34
André Nicolas
451k36422806
answered Mar 31 '15 at 0:34
André Nicolas
451k36422806
451k36422806
Yes, thank you. For the sake of familiarity I had used $h$, then decided not to, but did not manage to change all of them.
– André Nicolas
Mar 31 '15 at 0:41
Thanks for your reply - in any case, nice answer!
– user84413
Mar 31 '15 at 0:42
You are welcome. I consider the Taylor series approach the natural one, because it directly addresses the behaviour of the functions near $0$. But because of the order in which such things are done in standard North American calculus courses, Taylor series might be dealt with after this kind of limit problem.
– André Nicolas
Mar 31 '15 at 0:46
What is the property that makes this particular and other limits in general "fliparoveble"
– Paulo Martins
Mar 31 '15 at 1:16
They basically all are. We have $lim{xto a}f(x)=Lne 0$ if and only if $lim_{xto a}frac{1}{f(x)}=frac{1}{L}$. And we can even say useful things when $L= 0$, but $Lne 0$ in this problem.
– André Nicolas
Mar 31 '15 at 2:16
|
show 1 more comment
Yes, thank you. For the sake of familiarity I had used $h$, then decided not to, but did not manage to change all of them.
– André Nicolas
Mar 31 '15 at 0:41
Thanks for your reply - in any case, nice answer!
– user84413
Mar 31 '15 at 0:42
You are welcome. I consider the Taylor series approach the natural one, because it directly addresses the behaviour of the functions near $0$. But because of the order in which such things are done in standard North American calculus courses, Taylor series might be dealt with after this kind of limit problem.
– André Nicolas
Mar 31 '15 at 0:46
What is the property that makes this particular and other limits in general "fliparoveble"
– Paulo Martins
Mar 31 '15 at 1:16
They basically all are. We have $lim{xto a}f(x)=Lne 0$ if and only if $lim_{xto a}frac{1}{f(x)}=frac{1}{L}$. And we can even say useful things when $L= 0$, but $Lne 0$ in this problem.
– André Nicolas
Mar 31 '15 at 2:16
Yes, thank you. For the sake of familiarity I had used $h$, then decided not to, but did not manage to change all of them.
– André Nicolas
Mar 31 '15 at 0:41
Yes, thank you. For the sake of familiarity I had used $h$, then decided not to, but did not manage to change all of them.
– André Nicolas
Mar 31 '15 at 0:41
Thanks for your reply - in any case, nice answer!
– user84413
Mar 31 '15 at 0:42
Thanks for your reply - in any case, nice answer!
– user84413
Mar 31 '15 at 0:42
You are welcome. I consider the Taylor series approach the natural one, because it directly addresses the behaviour of the functions near $0$. But because of the order in which such things are done in standard North American calculus courses, Taylor series might be dealt with after this kind of limit problem.
– André Nicolas
Mar 31 '15 at 0:46
You are welcome. I consider the Taylor series approach the natural one, because it directly addresses the behaviour of the functions near $0$. But because of the order in which such things are done in standard North American calculus courses, Taylor series might be dealt with after this kind of limit problem.
– André Nicolas
Mar 31 '15 at 0:46
What is the property that makes this particular and other limits in general "fliparoveble"
– Paulo Martins
Mar 31 '15 at 1:16
What is the property that makes this particular and other limits in general "fliparoveble"
– Paulo Martins
Mar 31 '15 at 1:16
They basically all are. We have $lim{xto a}f(x)=Lne 0$ if and only if $lim_{xto a}frac{1}{f(x)}=frac{1}{L}$. And we can even say useful things when $L= 0$, but $Lne 0$ in this problem.
– André Nicolas
Mar 31 '15 at 2:16
They basically all are. We have $lim{xto a}f(x)=Lne 0$ if and only if $lim_{xto a}frac{1}{f(x)}=frac{1}{L}$. And we can even say useful things when $L= 0$, but $Lne 0$ in this problem.
– André Nicolas
Mar 31 '15 at 2:16
|
show 1 more comment
In the eternal words of Claude Leibovici, love Taylor Series.
We have
begin{align*}
ln(x^{2}+2x+4) &= ln(4) + frac{x}{2} + frac{x^{2}}{8} + mathcal{O}(x^{3})\
ln(x+4) &=ln(4) + frac{x}{4} -frac{x^{2}}{32} + mathcal{O}(x^{3}).
end{align*}
So
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)} = lim_{xto 0} frac{x}{ln(4) + frac{x}{2} + frac{x^{2}}{8} - left( ln(4) + frac{x}{4} -frac{x^{2}}{32}right) + mathcal{O}(x^{3})} = lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})}.$$
Divide the numerator and denominator by the lowest power of $x$ in the denominator to get
$$lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})} = lim_{xto 0} frac{frac{1}{x} x}{frac{1}{x}left( frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})right)} = lim_{xto 0}frac{1}{frac{1}{4} + frac{5x}{32} + mathcal{O}(x^{2})} = frac{1}{frac{1}{4}} = 4.$$
answered Mar 31 '15 at 0:38
JessicaK
4,46151733
In the final step of the Taylor series are you stating the inverse of the property lim x-> 00 of (a0x^n + a1x^n-1 ... an) = lim x->00 a0x^n? @JessicaK
– Paulo Martins
Mar 31 '15 at 1:00
@PauloMartins I am having some trouble understanding the equation that you wrote. I simply dropped the higher order terms of the polynomials because they do not affect the limit, but I edited my answer to avoid doing this.
– JessicaK
Mar 31 '15 at 1:03
@PauloMartins If "00" is supposed to be $infty$, hopefully by looking at the above modification, you should be able to convince yourself of an analogous property for limits of polynomials towards $0$ instead of $infty$.
– JessicaK
Mar 31 '15 at 1:13
Yes it was supposed to mean exactly that. And I have convinced myself of such property. Thanks
– Paulo Martins
Mar 31 '15 at 1:15
Thanks you, JessicaK for having noticed my love for Taylor ! Cheers :-)
– Claude Leibovici
Mar 31 '15 at 6:43
add a comment |
In the eternal words of Claude Leibovici, love Taylor Series.
We have
begin{align*}
ln(x^{2}+2x+4) &= ln(4) + frac{x}{2} + frac{x^{2}}{8} + mathcal{O}(x^{3})\
ln(x+4) &=ln(4) + frac{x}{4} -frac{x^{2}}{32} + mathcal{O}(x^{3}).
end{align*}
So
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)} = lim_{xto 0} frac{x}{ln(4) + frac{x}{2} + frac{x^{2}}{8} - left( ln(4) + frac{x}{4} -frac{x^{2}}{32}right) + mathcal{O}(x^{3})} = lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})}.$$
Divide the numerator and denominator by the lowest power of $x$ in the denominator to get
$$lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})} = lim_{xto 0} frac{frac{1}{x} x}{frac{1}{x}left( frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})right)} = lim_{xto 0}frac{1}{frac{1}{4} + frac{5x}{32} + mathcal{O}(x^{2})} = frac{1}{frac{1}{4}} = 4.$$
answered Mar 31 '15 at 0:38
JessicaK
4,46151733
In the final step of the Taylor series are you stating the inverse of the property lim x-> 00 of (a0x^n + a1x^n-1 ... an) = lim x->00 a0x^n? @JessicaK
– Paulo Martins
Mar 31 '15 at 1:00
@PauloMartins I am having some trouble understanding the equation that you wrote. I simply dropped the higher order terms of the polynomials because they do not affect the limit, but I edited my answer to avoid doing this.
– JessicaK
Mar 31 '15 at 1:03
@PauloMartins If "00" is supposed to be $infty$, hopefully by looking at the above modification, you should be able to convince yourself of an analogous property for limits of polynomials towards $0$ instead of $infty$.
– JessicaK
Mar 31 '15 at 1:13
Yes it was supposed to mean exactly that. And I have convinced myself of such property. Thanks
– Paulo Martins
Mar 31 '15 at 1:15
Thanks you, JessicaK for having noticed my love for Taylor ! Cheers :-)
– Claude Leibovici
Mar 31 '15 at 6:43
add a comment |
In the eternal words of Claude Leibovici, love Taylor Series.
We have
begin{align*}
ln(x^{2}+2x+4) &= ln(4) + frac{x}{2} + frac{x^{2}}{8} + mathcal{O}(x^{3})\
ln(x+4) &=ln(4) + frac{x}{4} -frac{x^{2}}{32} + mathcal{O}(x^{3}).
end{align*}
So
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)} = lim_{xto 0} frac{x}{ln(4) + frac{x}{2} + frac{x^{2}}{8} - left( ln(4) + frac{x}{4} -frac{x^{2}}{32}right) + mathcal{O}(x^{3})} = lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})}.$$
Divide the numerator and denominator by the lowest power of $x$ in the denominator to get
$$lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})} = lim_{xto 0} frac{frac{1}{x} x}{frac{1}{x}left( frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})right)} = lim_{xto 0}frac{1}{frac{1}{4} + frac{5x}{32} + mathcal{O}(x^{2})} = frac{1}{frac{1}{4}} = 4.$$
answered Mar 31 '15 at 0:38
JessicaK
4,46151733
In the eternal words of Claude Leibovici, love Taylor Series.
We have
begin{align*}
ln(x^{2}+2x+4) &= ln(4) + frac{x}{2} + frac{x^{2}}{8} + mathcal{O}(x^{3})\
ln(x+4) &=ln(4) + frac{x}{4} -frac{x^{2}}{32} + mathcal{O}(x^{3}).
end{align*}
So
$$lim_{xto 0} {xover ln (x^2+2x+4) - ln(x+4)} = lim_{xto 0} frac{x}{ln(4) + frac{x}{2} + frac{x^{2}}{8} - left( ln(4) + frac{x}{4} -frac{x^{2}}{32}right) + mathcal{O}(x^{3})} = lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})}.$$
Divide the numerator and denominator by the lowest power of $x$ in the denominator to get
$$lim_{xto 0} frac{x}{frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})} = lim_{xto 0} frac{frac{1}{x} x}{frac{1}{x}left( frac{x}{4} + frac{5x^{2}}{32} + mathcal{O}(x^{3})right)} = lim_{xto 0}frac{1}{frac{1}{4} + frac{5x}{32} + mathcal{O}(x^{2})} = frac{1}{frac{1}{4}} = 4.$$
answered Mar 31 '15 at 0:38
JessicaK
4,46151733
edited Mar 31 '15 at 1:02
answered Mar 31 '15 at 0:38
JessicaK
4,46151733
answered Mar 31 '15 at 0:38
JessicaK
4,46151733
answered Mar 31 '15 at 0:38
JessicaK
4,46151733
4,46151733
In the final step of the Taylor series are you stating the inverse of the property lim x-> 00 of (a0x^n + a1x^n-1 ... an) = lim x->00 a0x^n? @JessicaK
– Paulo Martins
Mar 31 '15 at 1:00
@PauloMartins I am having some trouble understanding the equation that you wrote. I simply dropped the higher order terms of the polynomials because they do not affect the limit, but I edited my answer to avoid doing this.
– JessicaK
Mar 31 '15 at 1:03
@PauloMartins If "00" is supposed to be $infty$, hopefully by looking at the above modification, you should be able to convince yourself of an analogous property for limits of polynomials towards $0$ instead of $infty$.
– JessicaK
Mar 31 '15 at 1:13
Yes it was supposed to mean exactly that. And I have convinced myself of such property. Thanks
– Paulo Martins
Mar 31 '15 at 1:15
Thanks you, JessicaK for having noticed my love for Taylor ! Cheers :-)
– Claude Leibovici
Mar 31 '15 at 6:43
add a comment |
In the final step of the Taylor series are you stating the inverse of the property lim x-> 00 of (a0x^n + a1x^n-1 ... an) = lim x->00 a0x^n? @JessicaK
– Paulo Martins
Mar 31 '15 at 1:00
@PauloMartins I am having some trouble understanding the equation that you wrote. I simply dropped the higher order terms of the polynomials because they do not affect the limit, but I edited my answer to avoid doing this.
– JessicaK
Mar 31 '15 at 1:03
@PauloMartins If "00" is supposed to be $infty$, hopefully by looking at the above modification, you should be able to convince yourself of an analogous property for limits of polynomials towards $0$ instead of $infty$.
– JessicaK
Mar 31 '15 at 1:13
Yes it was supposed to mean exactly that. And I have convinced myself of such property. Thanks
– Paulo Martins
Mar 31 '15 at 1:15
Thanks you, JessicaK for having noticed my love for Taylor ! Cheers :-)
– Claude Leibovici
Mar 31 '15 at 6:43
In the final step of the Taylor series are you stating the inverse of the property lim x-> 00 of (a0x^n + a1x^n-1 ... an) = lim x->00 a0x^n? @JessicaK
– Paulo Martins
Mar 31 '15 at 1:00
In the final step of the Taylor series are you stating the inverse of the property lim x-> 00 of (a0x^n + a1x^n-1 ... an) = lim x->00 a0x^n? @JessicaK
– Paulo Martins
Mar 31 '15 at 1:00
@PauloMartins I am having some trouble understanding the equation that you wrote. I simply dropped the higher order terms of the polynomials because they do not affect the limit, but I edited my answer to avoid doing this.
– JessicaK
Mar 31 '15 at 1:03
@PauloMartins I am having some trouble understanding the equation that you wrote. I simply dropped the higher order terms of the polynomials because they do not affect the limit, but I edited my answer to avoid doing this.
– JessicaK
Mar 31 '15 at 1:03
@PauloMartins If "00" is supposed to be $infty$, hopefully by looking at the above modification, you should be able to convince yourself of an analogous property for limits of polynomials towards $0$ instead of $infty$.
– JessicaK
Mar 31 '15 at 1:13
@PauloMartins If "00" is supposed to be $infty$, hopefully by looking at the above modification, you should be able to convince yourself of an analogous property for limits of polynomials towards $0$ instead of $infty$.
– JessicaK
Mar 31 '15 at 1:13
Yes it was supposed to mean exactly that. And I have convinced myself of such property. Thanks
– Paulo Martins
Mar 31 '15 at 1:15
Yes it was supposed to mean exactly that. And I have convinced myself of such property. Thanks
– Paulo Martins
Mar 31 '15 at 1:15
Thanks you, JessicaK for having noticed my love for Taylor ! Cheers :-)
– Claude Leibovici
Mar 31 '15 at 6:43
Thanks you, JessicaK for having noticed my love for Taylor ! Cheers :-)
– Claude Leibovici
Mar 31 '15 at 6:43
add a comment |
Rewrite it as
$$limlimits_{xrightarrow0}frac{x}{ln left( frac{x^2+2x+4}{x+4}right)}$$
Apply L'Hopital's rule:
$$limlimits_{xrightarrow0}frac{1}{frac{x^2+8x+4}{(x+4)(x^2+2x+4)}}$$
Then simply evaluate the limit:
$$frac{1}{frac{0^2+8times0+4}{(0+4)(0^2+2times0+4)}} = frac{1}{frac{4}{16}} = 4$$
answered Mar 31 '15 at 0:14
TokenToucan
2,4171419
Why not apply L'Hopital's rule with the difference of two logarithms rather than making it into a logarithm of a quotient first. That way you don't need the quotient rule. ${}qquad{}$
– Michael Hardy
Mar 31 '15 at 1:32
add a comment |
Rewrite it as
$$limlimits_{xrightarrow0}frac{x}{ln left( frac{x^2+2x+4}{x+4}right)}$$
Apply L'Hopital's rule:
$$limlimits_{xrightarrow0}frac{1}{frac{x^2+8x+4}{(x+4)(x^2+2x+4)}}$$
Then simply evaluate the limit:
$$frac{1}{frac{0^2+8times0+4}{(0+4)(0^2+2times0+4)}} = frac{1}{frac{4}{16}} = 4$$
answered Mar 31 '15 at 0:14
TokenToucan
2,4171419
Why not apply L'Hopital's rule with the difference of two logarithms rather than making it into a logarithm of a quotient first. That way you don't need the quotient rule. ${}qquad{}$
– Michael Hardy
Mar 31 '15 at 1:32
add a comment |
Rewrite it as
$$limlimits_{xrightarrow0}frac{x}{ln left( frac{x^2+2x+4}{x+4}right)}$$
Apply L'Hopital's rule:
$$limlimits_{xrightarrow0}frac{1}{frac{x^2+8x+4}{(x+4)(x^2+2x+4)}}$$
Then simply evaluate the limit:
$$frac{1}{frac{0^2+8times0+4}{(0+4)(0^2+2times0+4)}} = frac{1}{frac{4}{16}} = 4$$
answered Mar 31 '15 at 0:14
TokenToucan
2,4171419
Rewrite it as
$$limlimits_{xrightarrow0}frac{x}{ln left( frac{x^2+2x+4}{x+4}right)}$$
Apply L'Hopital's rule:
$$limlimits_{xrightarrow0}frac{1}{frac{x^2+8x+4}{(x+4)(x^2+2x+4)}}$$
Then simply evaluate the limit:
$$frac{1}{frac{0^2+8times0+4}{(0+4)(0^2+2times0+4)}} = frac{1}{frac{4}{16}} = 4$$
answered Mar 31 '15 at 0:14
TokenToucan
2,4171419
answered Mar 31 '15 at 0:14
TokenToucan
2,4171419
answered Mar 31 '15 at 0:14
TokenToucan
2,4171419
answered Mar 31 '15 at 0:14
TokenToucan
2,4171419
2,4171419
Why not apply L'Hopital's rule with the difference of two logarithms rather than making it into a logarithm of a quotient first. That way you don't need the quotient rule. ${}qquad{}$
– Michael Hardy
Mar 31 '15 at 1:32
add a comment |
Why not apply L'Hopital's rule with the difference of two logarithms rather than making it into a logarithm of a quotient first. That way you don't need the quotient rule. ${}qquad{}$
– Michael Hardy
Mar 31 '15 at 1:32
Why not apply L'Hopital's rule with the difference of two logarithms rather than making it into a logarithm of a quotient first. That way you don't need the quotient rule. ${}qquad{}$
– Michael Hardy
Mar 31 '15 at 1:32
Why not apply L'Hopital's rule with the difference of two logarithms rather than making it into a logarithm of a quotient first. That way you don't need the quotient rule. ${}qquad{}$
– Michael Hardy
Mar 31 '15 at 1:32
add a comment |
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