L'Hôpital's rule - How solve this limit question
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How to solve this ?
$$lim_{xto 0} f(x);text{where};f(x)=frac{ arctan(2x)}{ln (x)}$$
The answer is $0$. My question is when we plug in $0$ in $f(x)$, we get the form $frac{0}{infty}$, which is not an indeterminate form, so we might just write $0$ as answer directly OR if we apply L'Hôpital's rule, we would still get an answer as $0$. Which method is correct?
limits
edited yesterday
Chinnapparaj R
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asked yesterday
Amogh Joshi
183
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add a comment |
up vote
0
down vote
favorite
How to solve this ?
$$lim_{xto 0} f(x);text{where};f(x)=frac{ arctan(2x)}{ln (x)}$$
The answer is $0$. My question is when we plug in $0$ in $f(x)$, we get the form $frac{0}{infty}$, which is not an indeterminate form, so we might just write $0$ as answer directly OR if we apply L'Hôpital's rule, we would still get an answer as $0$. Which method is correct?
limits
edited yesterday
Chinnapparaj R
4,4101725
asked yesterday
Amogh Joshi
183
New contributor
Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We can't use L'Hôpital's rule for the reasons you mentioned, so that approach isn't correct. You are essentially right with the first explanation. To see this more clearly think of $f(x) = g(x)h(x)$, where $g(x) = arctan(2x)$, and $h(x) = 1/ln(x)$. Do you know how to do these two limits separately?
– Jabbath
yesterday
First one because L-Hopital's rule is only applicable when you get an indeterminate form. You can get different answers if the limits are not indeterminate.
– Sri Krishna Sahoo
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to solve this ?
$$lim_{xto 0} f(x);text{where};f(x)=frac{ arctan(2x)}{ln (x)}$$
The answer is $0$. My question is when we plug in $0$ in $f(x)$, we get the form $frac{0}{infty}$, which is not an indeterminate form, so we might just write $0$ as answer directly OR if we apply L'Hôpital's rule, we would still get an answer as $0$. Which method is correct?
limits
edited yesterday
Chinnapparaj R
4,4101725
asked yesterday
Amogh Joshi
183
New contributor
Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How to solve this ?
$$lim_{xto 0} f(x);text{where};f(x)=frac{ arctan(2x)}{ln (x)}$$
The answer is $0$. My question is when we plug in $0$ in $f(x)$, we get the form $frac{0}{infty}$, which is not an indeterminate form, so we might just write $0$ as answer directly OR if we apply L'Hôpital's rule, we would still get an answer as $0$. Which method is correct?
limits
limits
edited yesterday
Chinnapparaj R
4,4101725
asked yesterday
Amogh Joshi
183
New contributor
Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Chinnapparaj R
4,4101725
asked yesterday
Amogh Joshi
183
New contributor
Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Chinnapparaj R
4,4101725
edited yesterday
Chinnapparaj R
4,4101725
edited yesterday
Chinnapparaj R
4,4101725
4,4101725
asked yesterday
Amogh Joshi
183
New contributor
Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Amogh Joshi
183
asked yesterday
Amogh Joshi
183
183
New contributor
Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We can't use L'Hôpital's rule for the reasons you mentioned, so that approach isn't correct. You are essentially right with the first explanation. To see this more clearly think of $f(x) = g(x)h(x)$, where $g(x) = arctan(2x)$, and $h(x) = 1/ln(x)$. Do you know how to do these two limits separately?
– Jabbath
yesterday
First one because L-Hopital's rule is only applicable when you get an indeterminate form. You can get different answers if the limits are not indeterminate.
– Sri Krishna Sahoo
yesterday
add a comment |
We can't use L'Hôpital's rule for the reasons you mentioned, so that approach isn't correct. You are essentially right with the first explanation. To see this more clearly think of $f(x) = g(x)h(x)$, where $g(x) = arctan(2x)$, and $h(x) = 1/ln(x)$. Do you know how to do these two limits separately?
– Jabbath
yesterday
First one because L-Hopital's rule is only applicable when you get an indeterminate form. You can get different answers if the limits are not indeterminate.
– Sri Krishna Sahoo
yesterday
We can't use L'Hôpital's rule for the reasons you mentioned, so that approach isn't correct. You are essentially right with the first explanation. To see this more clearly think of $f(x) = g(x)h(x)$, where $g(x) = arctan(2x)$, and $h(x) = 1/ln(x)$. Do you know how to do these two limits separately?
– Jabbath
yesterday
We can't use L'Hôpital's rule for the reasons you mentioned, so that approach isn't correct. You are essentially right with the first explanation. To see this more clearly think of $f(x) = g(x)h(x)$, where $g(x) = arctan(2x)$, and $h(x) = 1/ln(x)$. Do you know how to do these two limits separately?
– Jabbath
yesterday
First one because L-Hopital's rule is only applicable when you get an indeterminate form. You can get different answers if the limits are not indeterminate.
– Sri Krishna Sahoo
yesterday
First one because L-Hopital's rule is only applicable when you get an indeterminate form. You can get different answers if the limits are not indeterminate.
– Sri Krishna Sahoo
yesterday
add a comment |
2 Answers
2
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1
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First of all, the limit cannot be $xto 0$, it must be $xto 0^+$, because of the domain of ln(x).
Second thing is that the L-Hopital rule is not applicable in this case, because it applies only for $frac{0}{0} or frac{infty}{infty}$ form, so the first method is correct.
Hope it is helpful.
answered yesterday
Crazy for maths
4887
add a comment |
up vote
0
down vote
Note that we can only consider $lim_{xto 0^+} f(x)$ and that we have a $frac 0{-infty}$ form which is not indeterminate.
answered yesterday
gimusi
84.3k74292
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
First of all, the limit cannot be $xto 0$, it must be $xto 0^+$, because of the domain of ln(x).
Second thing is that the L-Hopital rule is not applicable in this case, because it applies only for $frac{0}{0} or frac{infty}{infty}$ form, so the first method is correct.
Hope it is helpful.
answered yesterday
Crazy for maths
4887
add a comment |
up vote
1
down vote
First of all, the limit cannot be $xto 0$, it must be $xto 0^+$, because of the domain of ln(x).
Second thing is that the L-Hopital rule is not applicable in this case, because it applies only for $frac{0}{0} or frac{infty}{infty}$ form, so the first method is correct.
Hope it is helpful.
answered yesterday
Crazy for maths
4887
add a comment |
up vote
1
down vote
up vote
1
down vote
First of all, the limit cannot be $xto 0$, it must be $xto 0^+$, because of the domain of ln(x).
Second thing is that the L-Hopital rule is not applicable in this case, because it applies only for $frac{0}{0} or frac{infty}{infty}$ form, so the first method is correct.
Hope it is helpful.
answered yesterday
Crazy for maths
4887
First of all, the limit cannot be $xto 0$, it must be $xto 0^+$, because of the domain of ln(x).
Second thing is that the L-Hopital rule is not applicable in this case, because it applies only for $frac{0}{0} or frac{infty}{infty}$ form, so the first method is correct.
Hope it is helpful.
answered yesterday
Crazy for maths
4887
answered yesterday
Crazy for maths
4887
answered yesterday
Crazy for maths
4887
answered yesterday
Crazy for maths
4887
4887
add a comment |
add a comment |
up vote
0
down vote
Note that we can only consider $lim_{xto 0^+} f(x)$ and that we have a $frac 0{-infty}$ form which is not indeterminate.
answered yesterday
gimusi
84.3k74292
add a comment |
up vote
0
down vote
Note that we can only consider $lim_{xto 0^+} f(x)$ and that we have a $frac 0{-infty}$ form which is not indeterminate.
answered yesterday
gimusi
84.3k74292
add a comment |
up vote
0
down vote
up vote
0
down vote
Note that we can only consider $lim_{xto 0^+} f(x)$ and that we have a $frac 0{-infty}$ form which is not indeterminate.
answered yesterday
gimusi
84.3k74292
Note that we can only consider $lim_{xto 0^+} f(x)$ and that we have a $frac 0{-infty}$ form which is not indeterminate.
answered yesterday
gimusi
84.3k74292
answered yesterday
gimusi
84.3k74292
answered yesterday
gimusi
84.3k74292
answered yesterday
gimusi
84.3k74292
84.3k74292
add a comment |
add a comment |
Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.
Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.
Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.
Amogh Joshi is a new contributor. Be nice, and check out our Code of Conduct.
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We can't use L'Hôpital's rule for the reasons you mentioned, so that approach isn't correct. You are essentially right with the first explanation. To see this more clearly think of $f(x) = g(x)h(x)$, where $g(x) = arctan(2x)$, and $h(x) = 1/ln(x)$. Do you know how to do these two limits separately?
– Jabbath
yesterday
First one because L-Hopital's rule is only applicable when you get an indeterminate form. You can get different answers if the limits are not indeterminate.
– Sri Krishna Sahoo
yesterday