What is the technique where you perform a “u-substitution” on a limit called?
$begingroup$
I say u-substitution for lack of a better term, it's somewhat similar to u-substitution in integrals but a bit different. For example, if you want to evaluate:
$
lim_{x to 0^+}x^{x^2}$
What you can do to solve this is: set $u = x^2$
Then solve the original limit (in this case $x$ going to zero from the right) with $u$ as the function:
$lim_{x to 0^+}u$
Which gives zero since $0^2 = 0$
And then solve the limit where $u$ approaches the result above (in this case zero) of the substituted function (in this case $x^{x^2} = x^u$ since $u = x^2$):
$lim_{u to 0^+}x^u$
And then you get $1$ since anything to the power of zero is $1$.
Is there a name for this technique? I'd like to learn more about it and practice it but all I get when I search for "limit substitution" is things about direct limit substitution, not this stuff.
Any help is appreciated.
calculus limits
edited Dec 11 '18 at 21:31
Andrei
12.4k21128
asked Dec 11 '18 at 21:28
James RonaldJames Ronald
1237
$endgroup$
add a comment |
$begingroup$
I say u-substitution for lack of a better term, it's somewhat similar to u-substitution in integrals but a bit different. For example, if you want to evaluate:
$
lim_{x to 0^+}x^{x^2}$
What you can do to solve this is: set $u = x^2$
Then solve the original limit (in this case $x$ going to zero from the right) with $u$ as the function:
$lim_{x to 0^+}u$
Which gives zero since $0^2 = 0$
And then solve the limit where $u$ approaches the result above (in this case zero) of the substituted function (in this case $x^{x^2} = x^u$ since $u = x^2$):
$lim_{u to 0^+}x^u$
And then you get $1$ since anything to the power of zero is $1$.
Is there a name for this technique? I'd like to learn more about it and practice it but all I get when I search for "limit substitution" is things about direct limit substitution, not this stuff.
Any help is appreciated.
calculus limits
edited Dec 11 '18 at 21:31
Andrei
12.4k21128
asked Dec 11 '18 at 21:28
James RonaldJames Ronald
1237
$endgroup$
add a comment |
$begingroup$
I say u-substitution for lack of a better term, it's somewhat similar to u-substitution in integrals but a bit different. For example, if you want to evaluate:
$
lim_{x to 0^+}x^{x^2}$
What you can do to solve this is: set $u = x^2$
Then solve the original limit (in this case $x$ going to zero from the right) with $u$ as the function:
$lim_{x to 0^+}u$
Which gives zero since $0^2 = 0$
And then solve the limit where $u$ approaches the result above (in this case zero) of the substituted function (in this case $x^{x^2} = x^u$ since $u = x^2$):
$lim_{u to 0^+}x^u$
And then you get $1$ since anything to the power of zero is $1$.
Is there a name for this technique? I'd like to learn more about it and practice it but all I get when I search for "limit substitution" is things about direct limit substitution, not this stuff.
Any help is appreciated.
calculus limits
edited Dec 11 '18 at 21:31
Andrei
12.4k21128
asked Dec 11 '18 at 21:28
James RonaldJames Ronald
1237
$endgroup$
I say u-substitution for lack of a better term, it's somewhat similar to u-substitution in integrals but a bit different. For example, if you want to evaluate:
$
lim_{x to 0^+}x^{x^2}$
What you can do to solve this is: set $u = x^2$
Then solve the original limit (in this case $x$ going to zero from the right) with $u$ as the function:
$lim_{x to 0^+}u$
Which gives zero since $0^2 = 0$
And then solve the limit where $u$ approaches the result above (in this case zero) of the substituted function (in this case $x^{x^2} = x^u$ since $u = x^2$):
$lim_{u to 0^+}x^u$
And then you get $1$ since anything to the power of zero is $1$.
Is there a name for this technique? I'd like to learn more about it and practice it but all I get when I search for "limit substitution" is things about direct limit substitution, not this stuff.
Any help is appreciated.
calculus limits
calculus limits
edited Dec 11 '18 at 21:31
Andrei
12.4k21128
asked Dec 11 '18 at 21:28
James RonaldJames Ronald
1237
edited Dec 11 '18 at 21:31
Andrei
12.4k21128
asked Dec 11 '18 at 21:28
James RonaldJames Ronald
1237
edited Dec 11 '18 at 21:31
Andrei
12.4k21128
edited Dec 11 '18 at 21:31
Andrei
12.4k21128
edited Dec 11 '18 at 21:31
Andrei
12.4k21128
12.4k21128
asked Dec 11 '18 at 21:28
James RonaldJames Ronald
1237
asked Dec 11 '18 at 21:28
James RonaldJames Ronald
1237
asked Dec 11 '18 at 21:28
James RonaldJames Ronald
1237
1237
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Why don't you calculate the logarithm of the limit first:
$$ln lim_{x to 0^+}x^{x^2}=
lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$
answered Dec 11 '18 at 21:39
AndreiAndrei
12.4k21128
$endgroup$
$begingroup$
Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
$endgroup$
– James Ronald
Dec 11 '18 at 21:45
2
$begingroup$
It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
$endgroup$
– Andrei
Dec 11 '18 at 21:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Why don't you calculate the logarithm of the limit first:
$$ln lim_{x to 0^+}x^{x^2}=
lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$
answered Dec 11 '18 at 21:39
AndreiAndrei
12.4k21128
$endgroup$
$begingroup$
Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
$endgroup$
– James Ronald
Dec 11 '18 at 21:45
2
$begingroup$
It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
$endgroup$
– Andrei
Dec 11 '18 at 21:51
add a comment |
$begingroup$
Why don't you calculate the logarithm of the limit first:
$$ln lim_{x to 0^+}x^{x^2}=
lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$
answered Dec 11 '18 at 21:39
AndreiAndrei
12.4k21128
$endgroup$
$begingroup$
Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
$endgroup$
– James Ronald
Dec 11 '18 at 21:45
2
$begingroup$
It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
$endgroup$
– Andrei
Dec 11 '18 at 21:51
add a comment |
$begingroup$
Why don't you calculate the logarithm of the limit first:
$$ln lim_{x to 0^+}x^{x^2}=
lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$
answered Dec 11 '18 at 21:39
AndreiAndrei
12.4k21128
$endgroup$
Why don't you calculate the logarithm of the limit first:
$$ln lim_{x to 0^+}x^{x^2}=
lim_{x to 0^+}ln(x^{x^2})=lim_{x to 0^+}x^2ln x$$
answered Dec 11 '18 at 21:39
AndreiAndrei
12.4k21128
answered Dec 11 '18 at 21:39
AndreiAndrei
12.4k21128
answered Dec 11 '18 at 21:39
AndreiAndrei
12.4k21128
answered Dec 11 '18 at 21:39
AndreiAndrei
12.4k21128
12.4k21128
$begingroup$
Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
$endgroup$
– James Ronald
Dec 11 '18 at 21:45
2
$begingroup$
It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
$endgroup$
– Andrei
Dec 11 '18 at 21:51
add a comment |
$begingroup$
Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
$endgroup$
– James Ronald
Dec 11 '18 at 21:45
2
$begingroup$
It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
$endgroup$
– Andrei
Dec 11 '18 at 21:51
$begingroup$
Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
$endgroup$
– James Ronald
Dec 11 '18 at 21:45
$begingroup$
Well that's another way to solve the problem yes, but I was wondering just about the way I solved the question, not the solution itself. What is the name of the technique I used to solve the question? Thanks for the response by the way!
$endgroup$
– James Ronald
Dec 11 '18 at 21:45
2
2
$begingroup$
It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
$endgroup$
– Andrei
Dec 11 '18 at 21:51
$begingroup$
It gives the right answer in this particular case, but it's not a valid method. Try to solve $lim_{xto 0} sin(2x)/x$ using the same method
$endgroup$
– Andrei
Dec 11 '18 at 21:51
add a comment |
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