Mixing
Limits with Taylor series around zero
$begingroup$
I had some problems with the following two limits, which are supposed to be calculated with Taylor series:
$$
lim_{xto 0^+}frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}quadmbox{and}quad
lim_{xto 0^+}frac{(1-log{x})^{sin{x^2}}}{(arctan{x})^{3/2}}.
$$
Although the numerators are quite simple to develop in series, I stopped when I noticed that both denominators are not derivable in $x=0$, that is, we should not use Taylor series in $x=0$ to evaluate this functions around $0$. I wonder if is possible to consider right derivatives only, and study the behaviour of the denominators in a right neighborhood of $0$.
Thank you in advance for your help!
limits taylor-expansion indeterminate-forms
asked Jan 21 at 19:00
Fabio OriFabio Ori
161
$endgroup$
add a comment |
$begingroup$
I had some problems with the following two limits, which are supposed to be calculated with Taylor series:
$$
lim_{xto 0^+}frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}quadmbox{and}quad
lim_{xto 0^+}frac{(1-log{x})^{sin{x^2}}}{(arctan{x})^{3/2}}.
$$
Although the numerators are quite simple to develop in series, I stopped when I noticed that both denominators are not derivable in $x=0$, that is, we should not use Taylor series in $x=0$ to evaluate this functions around $0$. I wonder if is possible to consider right derivatives only, and study the behaviour of the denominators in a right neighborhood of $0$.
Thank you in advance for your help!
limits taylor-expansion indeterminate-forms
asked Jan 21 at 19:00
Fabio OriFabio Ori
161
$endgroup$
$begingroup$
A good question (+1).
$endgroup$
– hamam_Abdallah
Jan 21 at 19:12
add a comment |
$begingroup$
I had some problems with the following two limits, which are supposed to be calculated with Taylor series:
$$
lim_{xto 0^+}frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}quadmbox{and}quad
lim_{xto 0^+}frac{(1-log{x})^{sin{x^2}}}{(arctan{x})^{3/2}}.
$$
Although the numerators are quite simple to develop in series, I stopped when I noticed that both denominators are not derivable in $x=0$, that is, we should not use Taylor series in $x=0$ to evaluate this functions around $0$. I wonder if is possible to consider right derivatives only, and study the behaviour of the denominators in a right neighborhood of $0$.
Thank you in advance for your help!
limits taylor-expansion indeterminate-forms
asked Jan 21 at 19:00
Fabio OriFabio Ori
161
$endgroup$
I had some problems with the following two limits, which are supposed to be calculated with Taylor series:
$$
lim_{xto 0^+}frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}quadmbox{and}quad
lim_{xto 0^+}frac{(1-log{x})^{sin{x^2}}}{(arctan{x})^{3/2}}.
$$
Although the numerators are quite simple to develop in series, I stopped when I noticed that both denominators are not derivable in $x=0$, that is, we should not use Taylor series in $x=0$ to evaluate this functions around $0$. I wonder if is possible to consider right derivatives only, and study the behaviour of the denominators in a right neighborhood of $0$.
Thank you in advance for your help!
limits taylor-expansion indeterminate-forms
limits taylor-expansion indeterminate-forms
asked Jan 21 at 19:00
Fabio OriFabio Ori
161
asked Jan 21 at 19:00
Fabio OriFabio Ori
161
asked Jan 21 at 19:00
Fabio OriFabio Ori
161
asked Jan 21 at 19:00
Fabio OriFabio Ori
161
asked Jan 21 at 19:00
Fabio OriFabio Ori
161
161
$begingroup$
A good question (+1).
$endgroup$
– hamam_Abdallah
Jan 21 at 19:12
add a comment |
$begingroup$
A good question (+1).
$endgroup$
– hamam_Abdallah
Jan 21 at 19:12
$begingroup$
A good question (+1).
$endgroup$
– hamam_Abdallah
Jan 21 at 19:12
$begingroup$
A good question (+1).
$endgroup$
– hamam_Abdallah
Jan 21 at 19:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint. You can avoid the problem that $sqrt{x}$ is not differentiable at $0$ by considering just the Taylor series of $e^t=1+t+o(t)$ and $sin(t)=t+o(t)$ as $tto 0$. Then, as $xto 0^+$,
$$frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}=frac{(1+ sqrt{x}+o(sqrt{x}))-(1- sqrt{x}+o(sqrt{x}))}{sqrt{2x+o(x)}}=frac{sqrt{x}(2+o(1))}{sqrt{x}sqrt{2+o(1)}}.$$
Can you take it from here?
Use a similar approach also for the second one.
answered Jan 21 at 19:06
Robert ZRobert Z
99.9k1068140
$endgroup$
$begingroup$
Don't you want big O notation?
$endgroup$
– bounceback
Jan 21 at 19:08
$begingroup$
Little-o notation suffices.
$endgroup$
– Robert Z
Jan 21 at 19:09
$begingroup$
Oh, I see what you've done now, yes
$endgroup$
– bounceback
Jan 21 at 19:13
$begingroup$
Thank you @RobertZ, that is what I needed. For the second limit, I have just figured out that the numerator is $1+o(1)$, so it is not necessary to develop the denominator.
$endgroup$
– Fabio Ori
Jan 21 at 20:10
add a comment |
$begingroup$
Hint:
$$dfrac{e^{sqrt x}-e^{-sqrt x}}{sqrt{sin2x}}=dfrac{dfrac{e^{sqrt x}-1}{sqrt x}+dfrac{e^{-sqrt x}-1}{-sqrt x}}{sqrt2cdotsqrt{dfrac{sin2x}{2x}}}$$
answered Jan 21 at 19:09
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Thank you! This is an elegant solution, but, if I am not mistaking, it does not make usage of Taylor series.
$endgroup$
– Fabio Ori
Jan 21 at 19:29
add a comment |
$begingroup$
hint for the first
We have
$$e^X=1+X(1+epsilon(X))$$
then
$$e^{sqrt{x}}=1+sqrt{x}(1+epsilon(x))$$
and
when $xto 0^+$,
$$sqrt{sin(2x)}sim sqrt{2x}$$
thus your limit is
$$lim_{xto 0^+}frac{2sqrt{x}(1+epsilon(x))}{sqrt{2x}}=sqrt{2}.$$
TOMORROW I WILL LOSE POINTS .
NO ONE KNOWS THE REASON.
For this, i leave.
answered Jan 21 at 19:07
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint. You can avoid the problem that $sqrt{x}$ is not differentiable at $0$ by considering just the Taylor series of $e^t=1+t+o(t)$ and $sin(t)=t+o(t)$ as $tto 0$. Then, as $xto 0^+$,
$$frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}=frac{(1+ sqrt{x}+o(sqrt{x}))-(1- sqrt{x}+o(sqrt{x}))}{sqrt{2x+o(x)}}=frac{sqrt{x}(2+o(1))}{sqrt{x}sqrt{2+o(1)}}.$$
Can you take it from here?
Use a similar approach also for the second one.
answered Jan 21 at 19:06
Robert ZRobert Z
99.9k1068140
$endgroup$
$begingroup$
Don't you want big O notation?
$endgroup$
– bounceback
Jan 21 at 19:08
$begingroup$
Little-o notation suffices.
$endgroup$
– Robert Z
Jan 21 at 19:09
$begingroup$
Oh, I see what you've done now, yes
$endgroup$
– bounceback
Jan 21 at 19:13
$begingroup$
Thank you @RobertZ, that is what I needed. For the second limit, I have just figured out that the numerator is $1+o(1)$, so it is not necessary to develop the denominator.
$endgroup$
– Fabio Ori
Jan 21 at 20:10
add a comment |
$begingroup$
Hint. You can avoid the problem that $sqrt{x}$ is not differentiable at $0$ by considering just the Taylor series of $e^t=1+t+o(t)$ and $sin(t)=t+o(t)$ as $tto 0$. Then, as $xto 0^+$,
$$frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}=frac{(1+ sqrt{x}+o(sqrt{x}))-(1- sqrt{x}+o(sqrt{x}))}{sqrt{2x+o(x)}}=frac{sqrt{x}(2+o(1))}{sqrt{x}sqrt{2+o(1)}}.$$
Can you take it from here?
Use a similar approach also for the second one.
answered Jan 21 at 19:06
Robert ZRobert Z
99.9k1068140
$endgroup$
$begingroup$
Don't you want big O notation?
$endgroup$
– bounceback
Jan 21 at 19:08
$begingroup$
Little-o notation suffices.
$endgroup$
– Robert Z
Jan 21 at 19:09
$begingroup$
Oh, I see what you've done now, yes
$endgroup$
– bounceback
Jan 21 at 19:13
$begingroup$
Thank you @RobertZ, that is what I needed. For the second limit, I have just figured out that the numerator is $1+o(1)$, so it is not necessary to develop the denominator.
$endgroup$
– Fabio Ori
Jan 21 at 20:10
add a comment |
$begingroup$
Hint. You can avoid the problem that $sqrt{x}$ is not differentiable at $0$ by considering just the Taylor series of $e^t=1+t+o(t)$ and $sin(t)=t+o(t)$ as $tto 0$. Then, as $xto 0^+$,
$$frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}=frac{(1+ sqrt{x}+o(sqrt{x}))-(1- sqrt{x}+o(sqrt{x}))}{sqrt{2x+o(x)}}=frac{sqrt{x}(2+o(1))}{sqrt{x}sqrt{2+o(1)}}.$$
Can you take it from here?
Use a similar approach also for the second one.
answered Jan 21 at 19:06
Robert ZRobert Z
99.9k1068140
$endgroup$
Hint. You can avoid the problem that $sqrt{x}$ is not differentiable at $0$ by considering just the Taylor series of $e^t=1+t+o(t)$ and $sin(t)=t+o(t)$ as $tto 0$. Then, as $xto 0^+$,
$$frac{e^sqrt{x}-e^{-sqrt{x}}}{sqrt{sin{2x}}}=frac{(1+ sqrt{x}+o(sqrt{x}))-(1- sqrt{x}+o(sqrt{x}))}{sqrt{2x+o(x)}}=frac{sqrt{x}(2+o(1))}{sqrt{x}sqrt{2+o(1)}}.$$
Can you take it from here?
Use a similar approach also for the second one.
answered Jan 21 at 19:06
Robert ZRobert Z
99.9k1068140
edited Jan 21 at 19:08
answered Jan 21 at 19:06
Robert ZRobert Z
99.9k1068140
answered Jan 21 at 19:06
Robert ZRobert Z
99.9k1068140
answered Jan 21 at 19:06
Robert ZRobert Z
99.9k1068140
99.9k1068140
$begingroup$
Don't you want big O notation?
$endgroup$
– bounceback
Jan 21 at 19:08
$begingroup$
Little-o notation suffices.
$endgroup$
– Robert Z
Jan 21 at 19:09
$begingroup$
Oh, I see what you've done now, yes
$endgroup$
– bounceback
Jan 21 at 19:13
$begingroup$
Thank you @RobertZ, that is what I needed. For the second limit, I have just figured out that the numerator is $1+o(1)$, so it is not necessary to develop the denominator.
$endgroup$
– Fabio Ori
Jan 21 at 20:10
add a comment |
$begingroup$
Don't you want big O notation?
$endgroup$
– bounceback
Jan 21 at 19:08
$begingroup$
Little-o notation suffices.
$endgroup$
– Robert Z
Jan 21 at 19:09
$begingroup$
Oh, I see what you've done now, yes
$endgroup$
– bounceback
Jan 21 at 19:13
$begingroup$
Thank you @RobertZ, that is what I needed. For the second limit, I have just figured out that the numerator is $1+o(1)$, so it is not necessary to develop the denominator.
$endgroup$
– Fabio Ori
Jan 21 at 20:10
$begingroup$
Don't you want big O notation?
$endgroup$
– bounceback
Jan 21 at 19:08
$begingroup$
Don't you want big O notation?
$endgroup$
– bounceback
Jan 21 at 19:08
$begingroup$
Little-o notation suffices.
$endgroup$
– Robert Z
Jan 21 at 19:09
$begingroup$
Little-o notation suffices.
$endgroup$
– Robert Z
Jan 21 at 19:09
$begingroup$
Oh, I see what you've done now, yes
$endgroup$
– bounceback
Jan 21 at 19:13
$begingroup$
Oh, I see what you've done now, yes
$endgroup$
– bounceback
Jan 21 at 19:13
$begingroup$
Thank you @RobertZ, that is what I needed. For the second limit, I have just figured out that the numerator is $1+o(1)$, so it is not necessary to develop the denominator.
$endgroup$
– Fabio Ori
Jan 21 at 20:10
$begingroup$
Thank you @RobertZ, that is what I needed. For the second limit, I have just figured out that the numerator is $1+o(1)$, so it is not necessary to develop the denominator.
$endgroup$
– Fabio Ori
Jan 21 at 20:10
add a comment |
$begingroup$
Hint:
$$dfrac{e^{sqrt x}-e^{-sqrt x}}{sqrt{sin2x}}=dfrac{dfrac{e^{sqrt x}-1}{sqrt x}+dfrac{e^{-sqrt x}-1}{-sqrt x}}{sqrt2cdotsqrt{dfrac{sin2x}{2x}}}$$
answered Jan 21 at 19:09
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Thank you! This is an elegant solution, but, if I am not mistaking, it does not make usage of Taylor series.
$endgroup$
– Fabio Ori
Jan 21 at 19:29
add a comment |
$begingroup$
Hint:
$$dfrac{e^{sqrt x}-e^{-sqrt x}}{sqrt{sin2x}}=dfrac{dfrac{e^{sqrt x}-1}{sqrt x}+dfrac{e^{-sqrt x}-1}{-sqrt x}}{sqrt2cdotsqrt{dfrac{sin2x}{2x}}}$$
answered Jan 21 at 19:09
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
Thank you! This is an elegant solution, but, if I am not mistaking, it does not make usage of Taylor series.
$endgroup$
– Fabio Ori
Jan 21 at 19:29
add a comment |
$begingroup$
Hint:
$$dfrac{e^{sqrt x}-e^{-sqrt x}}{sqrt{sin2x}}=dfrac{dfrac{e^{sqrt x}-1}{sqrt x}+dfrac{e^{-sqrt x}-1}{-sqrt x}}{sqrt2cdotsqrt{dfrac{sin2x}{2x}}}$$
answered Jan 21 at 19:09
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
$$dfrac{e^{sqrt x}-e^{-sqrt x}}{sqrt{sin2x}}=dfrac{dfrac{e^{sqrt x}-1}{sqrt x}+dfrac{e^{-sqrt x}-1}{-sqrt x}}{sqrt2cdotsqrt{dfrac{sin2x}{2x}}}$$
answered Jan 21 at 19:09
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 21 at 19:09
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 21 at 19:09
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 21 at 19:09
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
Thank you! This is an elegant solution, but, if I am not mistaking, it does not make usage of Taylor series.
$endgroup$
– Fabio Ori
Jan 21 at 19:29
add a comment |
$begingroup$
Thank you! This is an elegant solution, but, if I am not mistaking, it does not make usage of Taylor series.
$endgroup$
– Fabio Ori
Jan 21 at 19:29
$begingroup$
Thank you! This is an elegant solution, but, if I am not mistaking, it does not make usage of Taylor series.
$endgroup$
– Fabio Ori
Jan 21 at 19:29
$begingroup$
Thank you! This is an elegant solution, but, if I am not mistaking, it does not make usage of Taylor series.
$endgroup$
– Fabio Ori
Jan 21 at 19:29
add a comment |
$begingroup$
hint for the first
We have
$$e^X=1+X(1+epsilon(X))$$
then
$$e^{sqrt{x}}=1+sqrt{x}(1+epsilon(x))$$
and
when $xto 0^+$,
$$sqrt{sin(2x)}sim sqrt{2x}$$
thus your limit is
$$lim_{xto 0^+}frac{2sqrt{x}(1+epsilon(x))}{sqrt{2x}}=sqrt{2}.$$
TOMORROW I WILL LOSE POINTS .
NO ONE KNOWS THE REASON.
For this, i leave.
answered Jan 21 at 19:07
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
hint for the first
We have
$$e^X=1+X(1+epsilon(X))$$
then
$$e^{sqrt{x}}=1+sqrt{x}(1+epsilon(x))$$
and
when $xto 0^+$,
$$sqrt{sin(2x)}sim sqrt{2x}$$
thus your limit is
$$lim_{xto 0^+}frac{2sqrt{x}(1+epsilon(x))}{sqrt{2x}}=sqrt{2}.$$
TOMORROW I WILL LOSE POINTS .
NO ONE KNOWS THE REASON.
For this, i leave.
answered Jan 21 at 19:07
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
hint for the first
We have
$$e^X=1+X(1+epsilon(X))$$
then
$$e^{sqrt{x}}=1+sqrt{x}(1+epsilon(x))$$
and
when $xto 0^+$,
$$sqrt{sin(2x)}sim sqrt{2x}$$
thus your limit is
$$lim_{xto 0^+}frac{2sqrt{x}(1+epsilon(x))}{sqrt{2x}}=sqrt{2}.$$
TOMORROW I WILL LOSE POINTS .
NO ONE KNOWS THE REASON.
For this, i leave.
answered Jan 21 at 19:07
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
hint for the first
We have
$$e^X=1+X(1+epsilon(X))$$
then
$$e^{sqrt{x}}=1+sqrt{x}(1+epsilon(x))$$
and
when $xto 0^+$,
$$sqrt{sin(2x)}sim sqrt{2x}$$
thus your limit is
$$lim_{xto 0^+}frac{2sqrt{x}(1+epsilon(x))}{sqrt{2x}}=sqrt{2}.$$
TOMORROW I WILL LOSE POINTS .
NO ONE KNOWS THE REASON.
For this, i leave.
answered Jan 21 at 19:07
hamam_Abdallahhamam_Abdallah
38.2k21634
edited Jan 21 at 19:15
answered Jan 21 at 19:07
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Jan 21 at 19:07
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Jan 21 at 19:07
hamam_Abdallahhamam_Abdallah
38.2k21634
38.2k21634
add a comment |
add a comment |
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$begingroup$
A good question (+1).
$endgroup$
– hamam_Abdallah
Jan 21 at 19:12