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Let $f:\mathbb{R} \to \mathbb{R}$ be such that $f''$ is continuous on $\mathbb{R}$ and $f(0)=1$ ,$f'(0)=0$ and $f''(0)=-1$ .

Then what is $\displaystyle\lim_{x\to \infty} \left[f\!\left(\sqrt{\frac{2}{x}}\,\right)\right]^x?$

When I was solving this problem, I supposed $f(x)$ to be a polynomial of degree two (because $f''$ is continuous) i.e. $f(x)=ax^2+bx+c$ and found coefficients with the help of given values . I got $f(x)=\frac{-x^2}{2} +1$. After solving , I found limit to be $e^{-1}$. I know this is a particular case.

Questions

$1$ : Will the limit be same for all functions with these properties ?

$2$ : Please give me some method which works for all such $f(x)$.

$3$: I want to practice more questions of this kind, please give me some references i.e. books, problem books, any online source.

Any kind of help will be highly appreciated. Thanks!

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5 Answers 5

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\begin{align*} &\lim_{x\rightarrow\infty}\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)^{x}\\ &=\lim_{x\rightarrow\infty}x\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)\\ &=\lim_{x\rightarrow\infty}\dfrac{\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)}{\dfrac{1}{x}}\\ &=\lim_{x\rightarrow\infty}\dfrac{\dfrac{1}{f\left(\sqrt{\dfrac{2}{x}}\right)}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}{-\dfrac{1}{x^{2}}}\\ &=\lim_{x\rightarrow\infty}\dfrac{\dfrac{1}{f\left(\sqrt{\dfrac{2}{x}}\right)}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)}{\sqrt{\dfrac{2}{x}}}\\ &=\lim_{x\rightarrow\infty}\dfrac{f'\left(\sqrt{\dfrac{2}{x}}\right)}{f\left(\sqrt{\dfrac{2}{x}}\right)\sqrt{\dfrac{2}{x}}}\\ &=\lim_{x\rightarrow\infty}\dfrac{f''\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}{\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}\cdot f\left(\sqrt{\dfrac{2}{x}}\right)+\sqrt{\dfrac{2}{x}}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}\\ &=\dfrac{-1}{1+0}\\ &=-1. \end{align*}

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  • $\begingroup$ I think you're almost there. Just need to get rid of the log, right? $\endgroup$ Commented Feb 23, 2018 at 16:19
  • $\begingroup$ Yes, you are right. $\endgroup$
    – user284331
    Commented Feb 23, 2018 at 16:35
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The problem is solved easily by taking logarithm. Let $2/x=t^2$ so that $t\to 0^{+}$. If $L$ is the desired limit then by continuity of logarithm $\log L$ is equal to the limit of the expression $$x\log f\left(\sqrt{\frac{2}{x}}\right)=\frac{2}{t^2}\cdot\underbrace{\frac{\log f(t)}{f(t)-1}}_{\to 1 } \cdot(f(t)-1)$$ which is equal to the limit of the expression $$\frac{f(t)-f(0)-tf'(0)}{t^2/2}$$ which is equal to $f''(0)=-1$ via Taylor's theorem (or a single application of L'Hospital's Rule). Thus $L=1/e$. Note that we only need existence of $f''(0)$ and not its continuity.


For practicing limit problems (or any math problems) this site is the best. You can ask as well as answer questions here. For a brief discussion of techniques of evaluation of limits you may refer to my blog series. As for books I always refer my favorite calculus book A Course of Pure Mathematics by G H Hardy. It should be available online for free.

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  • $\begingroup$ It means this limit will be same for all functions with these properties...? $\endgroup$
    – Learning
    Commented Feb 24, 2018 at 5:03
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    $\begingroup$ @PrithiviRaj: yes. You can see that I have not made any assumption about $f$ other than $f(0)=1,f'(0)=0,f''(0)=-1$. The limit is always $L=e^{f' '(0)}$. $\endgroup$
    – Paramanand Singh
    Commented Feb 24, 2018 at 5:05
  • $\begingroup$ I see , thanks a lot ! If you may answer my third question then I would be very grateful ,please! $\endgroup$
    – Learning
    Commented Feb 24, 2018 at 5:11
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    $\begingroup$ @PrithiviRaj: I have given some references now. $\endgroup$
    – Paramanand Singh
    Commented Feb 24, 2018 at 5:49
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This is a partial answer to your question and I'm not expecting any upvote, just to make it easier:

It seems for any polynomial with the degree equal or higher than $2$ this works fine: $f(x)=a_nx^n+a_{n-1}x^{n-1}+...-\dfrac{x^2}{2}+1$

It satisfies all conditions required by the question:

$f(0)=1, f'(0)=0, f''(0)=-1$

We may assume there are lots of function that are with this form as this can be represented by Taylor expansion of other functions when $n \to \infty$ .

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Hint: the limit is an indeterminate form $[1^\infty]$. Consider its logarithm $$ x \log \left( f \left( \sqrt{\frac{2}{x}} \right)\right), $$ assuming that $f>0$. Now you should ask yourself how to expand the argument of the logarithm as $x \to +\infty$. Of course, you will need an expansion of $f$ near zero.

Then you can try to compare the logarithm with $x$, and hope to get a conclusion.

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For $C^2$ functions, we have Taylor's theorem: $$ f(h)^x = [ f(0) + f'(0)h + f''(0)h^2/2 + o(h^2) ]^x, \quad (h\to 0)$$ Plugging in values gives \begin{align} f(\sqrt{2/x})^x &= \left[ 1 -x^{-1} + o(x^{-1})\right ]^x \\ & = \left[ 1 -x^{-1} \right ]^x\frac{\left[ 1 -x^{-1} + o(x^{-1})\right ]^x}{\left[ 1 -x^{-1} \right ]^x}\\ &=\left[ 1 -x^{-1} \right ]^x\left[ 1 -o(x^{-1}) \right ]^x, \quad (x\to\infty)\end{align} and $\lim_{x\to \infty}(1-o(x^{-1}))^x = 1$ so in the limit $x\to \infty$, the answer is $e^{-1}$. The linked proofs only show that $\lim_{x\to 0}(1-x^{-2}))^x = 1$ but most if not all of the proofs should generalise.

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