Let $\left(X_j\right)_{j\geq1}$ be i.i.d. with $E\{X_j\}=\mu$ and $\text{Var}\{X_j\}=\sigma^2$ (all $j$) with $0<\sigma^2<\infty$. Let $S_n=\sum\limits_{j=1}^{n}X_j$ and $Y_n=\dfrac{S_n-n\mu}{\sigma\sqrt{n}}$.
Let $\varphi_j$ be the characteristic function of $X_j-\mu$. Since the $\left(X_j\right)_{j\geq1}$ are i.i.d., $\varphi_j$ does not depend on $j$ and we write $\varphi$. One can show that $\varphi_{Y_n}(u)=\left(\varphi\left(\dfrac{u}{\sigma\sqrt{n}}\right)\right)^n$. Then, one can expand $\varphi$ in a Taylor expansion about $u=0$ to get $$\varphi(u)=1+0-\dfrac{\sigma^2u^2}{2}+u^2h(u)$$ with $h$ denoting the Peano remainder and $h(u)\rightarrow0$ as $u\rightarrow0$.
One can also show that $$\varphi_{Y_n}(u)=e^{n\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}$$ Taking limits as $n\rightarrow\infty$ and using for example L'Hopital rule, one gets $$\lim\limits_{n\to\infty}\varphi_{Y_n}(u)=e^{-\frac{u^2}{2}}$$

MY QUESTION: how could the above limit be solved by using L'Hopital's rule?

This is an indeterminate form $\infty\times0$. Thanks to the advice of user Mark Viola I get to the fact that limit can be rephrased as follows $$\lim\limits_{n\to\infty}\varphi_{Y_n}(u)=\lim\limits_{n\to\infty}e^{\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}}$$ At this point, we get an indeterminate form $\left[\dfrac{0}{0}\right]$. By continuity of exponential function $e$, we have $$\lim\limits_{n\to\infty}e^{\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}}=e^{\lim\limits_{n\to\infty}\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}}$$ At this point, focusing on the limit $$\lim\limits_{n\to\infty}\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}$$ I have tried to solve it by means of L'Hopital rule. $$\lim\limits_{n\to\infty}\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}\stackrel{H}=\lim\limits_{n\to\infty}\frac{\frac{d\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{dn}}{\frac{d\frac{1}{n}}{dn}}$$
I have some problem in computing $$\frac{d\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{dn}$$ Let $f(n)=\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)$. I know that $\dfrac{d\log\left(f(n)\right)}{dn}=\dfrac{f^{'}(n)}{f(n)}$ and my problems are related to the computation of $f^{'}(n)$ since I have no clue on how to compute the derivative of $\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)$ with respect to $n$, that is $$\dfrac{d \frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)}{dn} \tag{1}$$

Could you please explain to me in detail how to solve derivative $(1)$ so as to get to the final result $\lim\limits_{n\to\infty}\varphi_{Y_n}(u)=e^{-\frac{u^2}{2}}$?

  • $\begingroup$ You can't use this rule on sequences, they are not differentiable $\endgroup$
    – Andronicus
    Jan 13, 2020 at 17:09
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    $\begingroup$ I find it odd from the book to suggest using Taylor then L'Hospital. Here, as you saw, you get something quite ugly by using l'Hospital on the Taylor expansion. You also have to say something about $h'$, which is not obvious. Two things would feel more natural. 1) Use L'Hospital directly: $n \log \phi(u/\sigma \sqrt{n}) = \log \phi(ux/\sigma)/x^2$ with $x = 1/\sqrt{n}$, and L'Hospital. 2) Use Taylor all the way: $\log \phi(u/\sigma \sqrt{n}) = \log(1 - u^2/(2\sigma^2n) + o(1/n)) = - u^2/(2\sigma^2n) + o(1/n)$, and multiply by $n$ to get the result. $\endgroup$
    – Raoul
    Jun 6, 2020 at 23:52
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    $\begingroup$ Yes, $\phi''(0) = -\sigma^2$ indeed, thanks. The $u^2/\sigma^2$ comes from the chain rule applied twice. The derivative of $\log \phi(ux/\sigma)$ is $\frac{u}{\sigma} \phi'(ux/\sigma) / \phi(ux/\sigma)$, and then you get another $u/\sigma$ when you differentiate again. $\endgroup$
    – Raoul
    Jun 8, 2020 at 20:22
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    $\begingroup$ Thank you a lot, now it's all crystal-clear!!! Thanks for your time and patience. If you wish, you could summarize all your advice in an answer herebelow and I will be pleased to grant you the bounty which is open on this question, since you helped me significantly :) @Raoul $\endgroup$ Jun 8, 2020 at 20:30
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    $\begingroup$ Thanks, but no worries, happy it makes sense now :) $\endgroup$
    – Raoul
    Jun 8, 2020 at 22:49

1 Answer 1



Let $x=1/n$. Then examine the limit


Can you proceed now?

  • $\begingroup$ Pardon, when applying L'Hopital, computing the derivative of the argument of $\log$, that is $\dfrac{d\left(1-u^2x/2+(u^2/\sigma^2xh(u\sqrt{x}/\sigma))\right)}{dx}$, what do you get as result of derivative wrt $x$ of $(u^2/\sigma^2xh(u\sqrt{x}/\sigma))$? I think you get $0$, but I cannot figure out why one gets $0$. Could you please explain this to me? @MarkViola $\endgroup$ Jun 4, 2020 at 23:05
  • $\begingroup$ Take the derivative with respect to $x$. For $\log(f(x))$, its derivative is $\frac{f'(x)}{f(x)}$. So you should end up with $\frac{-u^2/2+o(x^{1/2})}{1-u^2x/2+o(x^{3/2})}$ $\endgroup$
    – Mark Viola
    Jun 4, 2020 at 23:34
  • $\begingroup$ In your answer, shouldn'it be $$\lim_{x\to0}\,\frac{\log\left(1-u^2x/2+(u^2x/\sigma^2h(u\sqrt{x}/\sigma))\right)}{x}$$ instead of $$\lim_{x\to0}\,\frac{\log\left(1-u^2x/2+(u^2/\sigma^2xh(u\sqrt{x}/\sigma))\right)}{x}$$? Sorry, I have noticed this right now. Additionally, how do you treat the function $h$ when computing derivative with respect to $x$? @MarkViola $\endgroup$ Jun 5, 2020 at 11:21
  • $\begingroup$ @Strictly_increasing Concerning your last comment, you both are right. Note that $a/bc = (a/b)c = (ac)/b$ by the usual bracketing convention. He did not write $a/(bc)$. $\endgroup$
    – J. De Ro
    Jun 5, 2020 at 12:54
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    $\begingroup$ @Strictly_increasing As $\varepsilon-\delta$ stated, $u^2x/\sigma^2=u^2/\sigma^2 x$. And $h(u\sqrt{x}/\sigma)\to 0$ as $x\to 0$. The notation implies that for any $\varepsilon>0$, $|h|\le \varepsilon \sqrt{x}$. $\endgroup$
    – Mark Viola
    Jun 5, 2020 at 13:22

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