L'Hôpital's rule - How solve this limit question
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 Nov 14 at 6:23
Chinnapparaj R
4,6081725
asked Nov 14 at 6:13
Amogh Joshi
183
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 Nov 14 at 6:23
Chinnapparaj R
4,6081725
asked Nov 14 at 6:13
Amogh Joshi
183
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
Nov 14 at 6:21
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
Nov 14 at 6:22
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 Nov 14 at 6:23
Chinnapparaj R
4,6081725
asked Nov 14 at 6:13
Amogh Joshi
183
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 Nov 14 at 6:23
Chinnapparaj R
4,6081725
asked Nov 14 at 6:13
Amogh Joshi
183
edited Nov 14 at 6:23
Chinnapparaj R
4,6081725
asked Nov 14 at 6:13
Amogh Joshi
183
edited Nov 14 at 6:23
Chinnapparaj R
4,6081725
edited Nov 14 at 6:23
Chinnapparaj R
4,6081725
edited Nov 14 at 6:23
Chinnapparaj R
4,6081725
4,6081725
asked Nov 14 at 6:13
Amogh Joshi
183
asked Nov 14 at 6:13
Amogh Joshi
183
asked Nov 14 at 6:13
Amogh Joshi
183
183
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
Nov 14 at 6:21
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
Nov 14 at 6:22
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
Nov 14 at 6:21
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
Nov 14 at 6:22
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
Nov 14 at 6:21
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
Nov 14 at 6:21
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
Nov 14 at 6:22
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
Nov 14 at 6:22
add a comment |
2 Answers
2
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 Nov 14 at 6:22
Crazy for maths
57210
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 Nov 14 at 8:07
gimusi
87.2k74393
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 Nov 14 at 6:22
Crazy for maths
57210
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 Nov 14 at 6:22
Crazy for maths
57210
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 Nov 14 at 6:22
Crazy for maths
57210
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 Nov 14 at 6:22
Crazy for maths
57210
answered Nov 14 at 6:22
Crazy for maths
57210
answered Nov 14 at 6:22
Crazy for maths
57210
answered Nov 14 at 6:22
Crazy for maths
57210
57210
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 Nov 14 at 8:07
gimusi
87.2k74393
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 Nov 14 at 8:07
gimusi
87.2k74393
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 Nov 14 at 8:07
gimusi
87.2k74393
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 Nov 14 at 8:07
gimusi
87.2k74393
answered Nov 14 at 8:07
gimusi
87.2k74393
answered Nov 14 at 8:07
gimusi
87.2k74393
answered Nov 14 at 8:07
gimusi
87.2k74393
87.2k74393
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997865%2flh%25c3%25b4pitals-rule-how-solve-this-limit-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
Nov 14 at 6:21
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
Nov 14 at 6:22