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Finding $lim_xto inftyfleft(sqrtfrac2xright)^x$, when $f(0)=1$, $f^prime(0)=0$, and $f^primeprime(0)=-1$
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Let $f:mathbb Rrightarrow mathbb R$ such that $f^primeprime$ is continuous on $R$ and $f(0)=1$, $f^prime(0)=0$, and $f^primeprime(0)=-1$ . Then, I have to find $$lim_xto inftyfleft(sqrtfrac2xright)^x$$
I can see that it is in indeterminate form $1^infty$ and have let $fleft(sqrtdfrac2xright)^x=t$ and taken $log$ on both sides but I am not able to differentiate the function.
calculus
edited Jul 23 at 6:03
Abcd
2,3361624
asked Jul 22 at 17:03
user416571
223
add a comment |Â
up vote
3
down vote
favorite
Let $f:mathbb Rrightarrow mathbb R$ such that $f^primeprime$ is continuous on $R$ and $f(0)=1$, $f^prime(0)=0$, and $f^primeprime(0)=-1$ . Then, I have to find $$lim_xto inftyfleft(sqrtfrac2xright)^x$$
I can see that it is in indeterminate form $1^infty$ and have let $fleft(sqrtdfrac2xright)^x=t$ and taken $log$ on both sides but I am not able to differentiate the function.
calculus
edited Jul 23 at 6:03
Abcd
2,3361624
asked Jul 22 at 17:03
user416571
223
Are you 100% sure about the question? The answer, as stated, is $+infty$, and the assumptions on $f''$ are not used for that. Wasn't it $f'(0)=0$ instead?
– Clement C.
Jul 22 at 17:15
sorry , i have made mistake .You are right.
– user416571
Jul 23 at 2:16
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Jul 25 at 20:22
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up vote
3
down vote
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up vote
3
down vote
favorite
Let $f:mathbb Rrightarrow mathbb R$ such that $f^primeprime$ is continuous on $R$ and $f(0)=1$, $f^prime(0)=0$, and $f^primeprime(0)=-1$ . Then, I have to find $$lim_xto inftyfleft(sqrtfrac2xright)^x$$
I can see that it is in indeterminate form $1^infty$ and have let $fleft(sqrtdfrac2xright)^x=t$ and taken $log$ on both sides but I am not able to differentiate the function.
calculus
edited Jul 23 at 6:03
Abcd
2,3361624
asked Jul 22 at 17:03
user416571
223
Let $f:mathbb Rrightarrow mathbb R$ such that $f^primeprime$ is continuous on $R$ and $f(0)=1$, $f^prime(0)=0$, and $f^primeprime(0)=-1$ . Then, I have to find $$lim_xto inftyfleft(sqrtfrac2xright)^x$$
I can see that it is in indeterminate form $1^infty$ and have let $fleft(sqrtdfrac2xright)^x=t$ and taken $log$ on both sides but I am not able to differentiate the function.
calculus
edited Jul 23 at 6:03
Abcd
2,3361624
asked Jul 22 at 17:03
user416571
223
edited Jul 23 at 6:03
Abcd
2,3361624
edited Jul 23 at 6:03
Abcd
2,3361624
edited Jul 23 at 6:03
Abcd
2,3361624
2,3361624
asked Jul 22 at 17:03
user416571
223
asked Jul 22 at 17:03
user416571
223
asked Jul 22 at 17:03
user416571
223
223
Are you 100% sure about the question? The answer, as stated, is $+infty$, and the assumptions on $f''$ are not used for that. Wasn't it $f'(0)=0$ instead?
– Clement C.
Jul 22 at 17:15
sorry , i have made mistake .You are right.
– user416571
Jul 23 at 2:16
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Jul 25 at 20:22
add a comment |Â
Are you 100% sure about the question? The answer, as stated, is $+infty$, and the assumptions on $f''$ are not used for that. Wasn't it $f'(0)=0$ instead?
– Clement C.
Jul 22 at 17:15
sorry , i have made mistake .You are right.
– user416571
Jul 23 at 2:16
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Jul 25 at 20:22
Are you 100% sure about the question? The answer, as stated, is $+infty$, and the assumptions on $f''$ are not used for that. Wasn't it $f'(0)=0$ instead?
– Clement C.
Jul 22 at 17:15
Are you 100% sure about the question? The answer, as stated, is $+infty$, and the assumptions on $f''$ are not used for that. Wasn't it $f'(0)=0$ instead?
– Clement C.
Jul 22 at 17:15
sorry , i have made mistake .You are right.
– user416571
Jul 23 at 2:16
sorry , i have made mistake .You are right.
– user416571
Jul 23 at 2:16
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Jul 25 at 20:22
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Jul 25 at 20:22
add a comment |Â
1 Answer
1
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First, note that since $f(0) > 0$ and $f$ is continuous, then $f(u)>0$ when $u$ is small enough. Thus the quantity $fleft(sqrtfrac2xright)^x$ is well-defined for $x$ large enough.
A key tool here is to use Taylor expansions, which tell you that "around $0$" (and under some regularity assumptions, which $f$ satisfies here) you can approximate $f(x)$ by the polynomial $f(0)+f'(0)x+frac12f''(0)x^2$. And polynomials are nice. Dealing with the case of
$
tildef(x) = 1-fracx^22
$
(instead of the general $f$ you have) is much easier (try is as a warmup!). This is essentially what we will be able to do.
Specifically, by Taylor's theorem (as indeed $f$ is twice differentiable at $0$) we have, using the Landau notation,
$$
f(u) = (0)+f'(0)u+frac12f''(0)u^2 + o(u^2)
= 1 - fracu^22 + o(u^2),. tag1
$$
Now, by a change of variable, letting $ustackrelrm def= sqrtfrac2x xrightarrow[xtoinfty]0^+$, we have
$$
fleft(sqrtfrac2xright)^x = e^x ln fleft(sqrtfrac2xright)
= e^frac2u^2 ln (1 - fracu^22 + o(u^2)) tag2
$$
At this point, using the standard fact that $ln(1+u) = u+o(u)$ when $uto0$ (to see this: it is equivalent to writing $lim_uto 0fracln(1+u)u=1$), we have by (2)
$$
fleft(sqrtfrac2xright)^x
= e^frac2u^2 ln (1 - fracu^22 + o(u^2))
= e^frac2u^2 (- fracu^22 + o(u^2))
= e^- 1 + o(1) tag3
$$
i.e.,
$$
lim_xtoinftyfleft(sqrtfrac2xright)^x
= lim_uto 0e^- 1 + o(1) = boxede^-1 tag4
$$
answered Jul 22 at 17:28
Clement C.
47.1k33682
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
First, note that since $f(0) > 0$ and $f$ is continuous, then $f(u)>0$ when $u$ is small enough. Thus the quantity $fleft(sqrtfrac2xright)^x$ is well-defined for $x$ large enough.
A key tool here is to use Taylor expansions, which tell you that "around $0$" (and under some regularity assumptions, which $f$ satisfies here) you can approximate $f(x)$ by the polynomial $f(0)+f'(0)x+frac12f''(0)x^2$. And polynomials are nice. Dealing with the case of
$
tildef(x) = 1-fracx^22
$
(instead of the general $f$ you have) is much easier (try is as a warmup!). This is essentially what we will be able to do.
Specifically, by Taylor's theorem (as indeed $f$ is twice differentiable at $0$) we have, using the Landau notation,
$$
f(u) = (0)+f'(0)u+frac12f''(0)u^2 + o(u^2)
= 1 - fracu^22 + o(u^2),. tag1
$$
Now, by a change of variable, letting $ustackrelrm def= sqrtfrac2x xrightarrow[xtoinfty]0^+$, we have
$$
fleft(sqrtfrac2xright)^x = e^x ln fleft(sqrtfrac2xright)
= e^frac2u^2 ln (1 - fracu^22 + o(u^2)) tag2
$$
At this point, using the standard fact that $ln(1+u) = u+o(u)$ when $uto0$ (to see this: it is equivalent to writing $lim_uto 0fracln(1+u)u=1$), we have by (2)
$$
fleft(sqrtfrac2xright)^x
= e^frac2u^2 ln (1 - fracu^22 + o(u^2))
= e^frac2u^2 (- fracu^22 + o(u^2))
= e^- 1 + o(1) tag3
$$
i.e.,
$$
lim_xtoinftyfleft(sqrtfrac2xright)^x
= lim_uto 0e^- 1 + o(1) = boxede^-1 tag4
$$
answered Jul 22 at 17:28
Clement C.
47.1k33682
add a comment |Â
up vote
5
down vote
First, note that since $f(0) > 0$ and $f$ is continuous, then $f(u)>0$ when $u$ is small enough. Thus the quantity $fleft(sqrtfrac2xright)^x$ is well-defined for $x$ large enough.
A key tool here is to use Taylor expansions, which tell you that "around $0$" (and under some regularity assumptions, which $f$ satisfies here) you can approximate $f(x)$ by the polynomial $f(0)+f'(0)x+frac12f''(0)x^2$. And polynomials are nice. Dealing with the case of
$
tildef(x) = 1-fracx^22
$
(instead of the general $f$ you have) is much easier (try is as a warmup!). This is essentially what we will be able to do.
Specifically, by Taylor's theorem (as indeed $f$ is twice differentiable at $0$) we have, using the Landau notation,
$$
f(u) = (0)+f'(0)u+frac12f''(0)u^2 + o(u^2)
= 1 - fracu^22 + o(u^2),. tag1
$$
Now, by a change of variable, letting $ustackrelrm def= sqrtfrac2x xrightarrow[xtoinfty]0^+$, we have
$$
fleft(sqrtfrac2xright)^x = e^x ln fleft(sqrtfrac2xright)
= e^frac2u^2 ln (1 - fracu^22 + o(u^2)) tag2
$$
At this point, using the standard fact that $ln(1+u) = u+o(u)$ when $uto0$ (to see this: it is equivalent to writing $lim_uto 0fracln(1+u)u=1$), we have by (2)
$$
fleft(sqrtfrac2xright)^x
= e^frac2u^2 ln (1 - fracu^22 + o(u^2))
= e^frac2u^2 (- fracu^22 + o(u^2))
= e^- 1 + o(1) tag3
$$
i.e.,
$$
lim_xtoinftyfleft(sqrtfrac2xright)^x
= lim_uto 0e^- 1 + o(1) = boxede^-1 tag4
$$
answered Jul 22 at 17:28
Clement C.
47.1k33682
add a comment |Â
up vote
5
down vote
up vote
5
down vote
First, note that since $f(0) > 0$ and $f$ is continuous, then $f(u)>0$ when $u$ is small enough. Thus the quantity $fleft(sqrtfrac2xright)^x$ is well-defined for $x$ large enough.
A key tool here is to use Taylor expansions, which tell you that "around $0$" (and under some regularity assumptions, which $f$ satisfies here) you can approximate $f(x)$ by the polynomial $f(0)+f'(0)x+frac12f''(0)x^2$. And polynomials are nice. Dealing with the case of
$
tildef(x) = 1-fracx^22
$
(instead of the general $f$ you have) is much easier (try is as a warmup!). This is essentially what we will be able to do.
Specifically, by Taylor's theorem (as indeed $f$ is twice differentiable at $0$) we have, using the Landau notation,
$$
f(u) = (0)+f'(0)u+frac12f''(0)u^2 + o(u^2)
= 1 - fracu^22 + o(u^2),. tag1
$$
Now, by a change of variable, letting $ustackrelrm def= sqrtfrac2x xrightarrow[xtoinfty]0^+$, we have
$$
fleft(sqrtfrac2xright)^x = e^x ln fleft(sqrtfrac2xright)
= e^frac2u^2 ln (1 - fracu^22 + o(u^2)) tag2
$$
At this point, using the standard fact that $ln(1+u) = u+o(u)$ when $uto0$ (to see this: it is equivalent to writing $lim_uto 0fracln(1+u)u=1$), we have by (2)
$$
fleft(sqrtfrac2xright)^x
= e^frac2u^2 ln (1 - fracu^22 + o(u^2))
= e^frac2u^2 (- fracu^22 + o(u^2))
= e^- 1 + o(1) tag3
$$
i.e.,
$$
lim_xtoinftyfleft(sqrtfrac2xright)^x
= lim_uto 0e^- 1 + o(1) = boxede^-1 tag4
$$
answered Jul 22 at 17:28
Clement C.
47.1k33682
First, note that since $f(0) > 0$ and $f$ is continuous, then $f(u)>0$ when $u$ is small enough. Thus the quantity $fleft(sqrtfrac2xright)^x$ is well-defined for $x$ large enough.
A key tool here is to use Taylor expansions, which tell you that "around $0$" (and under some regularity assumptions, which $f$ satisfies here) you can approximate $f(x)$ by the polynomial $f(0)+f'(0)x+frac12f''(0)x^2$. And polynomials are nice. Dealing with the case of
$
tildef(x) = 1-fracx^22
$
(instead of the general $f$ you have) is much easier (try is as a warmup!). This is essentially what we will be able to do.
Specifically, by Taylor's theorem (as indeed $f$ is twice differentiable at $0$) we have, using the Landau notation,
$$
f(u) = (0)+f'(0)u+frac12f''(0)u^2 + o(u^2)
= 1 - fracu^22 + o(u^2),. tag1
$$
Now, by a change of variable, letting $ustackrelrm def= sqrtfrac2x xrightarrow[xtoinfty]0^+$, we have
$$
fleft(sqrtfrac2xright)^x = e^x ln fleft(sqrtfrac2xright)
= e^frac2u^2 ln (1 - fracu^22 + o(u^2)) tag2
$$
At this point, using the standard fact that $ln(1+u) = u+o(u)$ when $uto0$ (to see this: it is equivalent to writing $lim_uto 0fracln(1+u)u=1$), we have by (2)
$$
fleft(sqrtfrac2xright)^x
= e^frac2u^2 ln (1 - fracu^22 + o(u^2))
= e^frac2u^2 (- fracu^22 + o(u^2))
= e^- 1 + o(1) tag3
$$
i.e.,
$$
lim_xtoinftyfleft(sqrtfrac2xright)^x
= lim_uto 0e^- 1 + o(1) = boxede^-1 tag4
$$
answered Jul 22 at 17:28
Clement C.
47.1k33682
edited Jul 23 at 6:06
answered Jul 22 at 17:28
Clement C.
47.1k33682
answered Jul 22 at 17:28
Clement C.
47.1k33682
answered Jul 22 at 17:28
Clement C.
47.1k33682
47.1k33682
add a comment |Â
add a comment |Â
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Are you 100% sure about the question? The answer, as stated, is $+infty$, and the assumptions on $f''$ are not used for that. Wasn't it $f'(0)=0$ instead?
– Clement C.
Jul 22 at 17:15
sorry , i have made mistake .You are right.
– user416571
Jul 23 at 2:16
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Jul 25 at 20:22