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Evaluate the following limit:

$$L=\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$$

Using $\ln(1+x)=x-x^2/2+x^3/3-\cdots$

I got $(1+x)^{1/x}=e^{1-x/2+x^2/3-\cdots}$

Could some tell me how to proceed further?

Behrouz Maleki
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MathGeek
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3 Answers3

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Indeed, the expansion of $(1+x)^{1/x}$ about $x=0$ is $e - \frac{ex}{2} + \frac{11e}{24} x^2 + O(x^3)$, so the limit is then $\dfrac{11e}{24}$.

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To get this expansion, start with

$$(1+x)^{1/x} = \exp \left({1 - x/2 + x^2 / 3 - \dots}\right)$$

which you already found and put this into the power series of $\exp$ to get

$$\sum_{n=0}^{\infty} \frac{(1 - x/2 + x^2/3 - \dots)^n}{n!}$$

The coefficient of $x^0$ is

$$\sum_{n=0}^{\infty} \frac{1}{n!} = e$$

The coefficient of $x^1$ in $\dfrac{(1-x/2+x^2/3 - \dots)^n}{n!}$ is $-\frac{n/2}{n!}$, hence the coefficient in total is

$$\sum_{n=1}^{\infty} -\frac{1}{2 (n-1)!} = -\frac{e}{2}$$

I'll leave the coefficient of $x^2$ to you. (It's the hardest, but not too bad).

MT_
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  • It would be better and much simpler to evaluate the coefficients if you write $$\exp\left(1 - \frac{x}{2} + \frac{x^{2}}{3} - \cdots\right) = e\cdot\exp\left(- \frac{x}{2} + \frac{x^{2}}{3} - \cdots\right)$$ and then you don't need to sum an infinite series to calculate coefficients. Thus coefficient of $x^{2}$ is $$e\left(\frac{1}{3} + \frac{1}{2}\cdot\left(\frac{1}{2}\right)^{2}\right) = \frac{11e}{24}$$ – Paramanand Singh Jul 19 '16 at 05:46
  • @ParamanandSingh Why is that expression the coefficient? I don't see where it follows from - I see that you took the coefficient of $x^2$ and added $1/2$ times the square of the coefficient of $x$, but I don't see why that is immediate when you manipulate the series like that – MT_ Jul 19 '16 at 05:48
  • we have $$\exp(-x/2 + x^{2}/3 + \cdots) = 1 + (-x/2 + x^{2}/3 - \cdots) + \frac{(-x/2 + x^{2}/3 - \cdots)^{2}}{2} + \cdots$$ and note that further terms involve $x^{3}$ and higher powers. So the coefficient of $x^{2}$ is $1/3 + (-1/2)^{2}\cdot (1/2) = 11/24$. Thus you don't need to sum infinite numbers of terms to obtain coefficients. You can similarly get coefficients of $x^{3}, x^{4}$ with far too less effort. – Paramanand Singh Jul 19 '16 at 05:51
  • @ParamanandSingh Oh, I see. Yeah, I suppose it's a bit simpler. – MT_ Jul 19 '16 at 05:57
  • Let $s = 1 - x/2 + x^{2}/3 - \cdots, t = -x/2 + x^{2}/3 - \cdots$. Now the term $x^{k}$ is present in each term of expansion $\exp(s) = 1 + s + s^{2}/2! + \cdots$ but $x^{k}$ is present in expansion $\exp(t) = 1 + t + t^{2}/2! + \cdots$ only till $t^{k}/k!$. – Paramanand Singh Jul 19 '16 at 05:57
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You can proceed in the following manner \begin{align} L &= \lim_{x \to 0}\dfrac{(1 + x)^{1/x} - e + \dfrac{ex}{2}}{x^{2}}\notag\\ &= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1 + \dfrac{x}{2}}{x^{2}}\notag\\ &= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - \exp\left(\log\left(1 - \dfrac{x}{2}\right)\right)}{x^{2}}\notag\\ &= e\lim_{x \to 0}\exp\left(\log\left(1 - \dfrac{x}{2}\right)\right)\cdot\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)\right) - 1}{x^{2}}\notag\\ &= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)\right) - 1}{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}\cdot\dfrac{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}{x^{2}}\notag\\ &= e\lim_{x \to 0}\dfrac{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}{x^{2}}\notag\\ &= e\lim_{x \to 0}\dfrac{\left(1 - \dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right) - 1 + \left(\dfrac{x}{2} + \dfrac{x^{2}}{8} + o(x^{2})\right)}{x^{2}}\text{ (using Taylor series)}\notag\\ &= e\left(\frac{1}{3} + \frac{1}{8}\right)\notag\\ &= \frac{11e}{24}\notag \end{align} The above approach uses the standard limit $$\lim_{t \to 0}\frac{\exp(t) - 1}{t} = 1$$ Taylor series is used only when necessary and this approach avoids any multiplication/division of infinite series.

Paramanand Singh
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1

It's a well-known fact that $$\log(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3+\underset{x\to 0}{o}(x^3)$$

Hence we get $$\dfrac 1x\log(1+x)=1+u(x)$$ where $$u(x)=-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)$$ Notice that $$\lim_{x\to 0} u(x)=0$$ We can write $$(1+x)^{1/x}=ee^{u(x)}$$

But we know that $$e^u=1+u+\dfrac{u^2}2+\underset{u\to 0}o(u^2)$$

Now using the identity $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$$ we quickly find, since we only need the terms whose degrees are less than three (the other terms are all some $\underset{x\to 0}o(x^2)$), that $$(u(x))^2=\left(-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)\right)^2=\dfrac{x^2}4+\underset{x\to 0}o(x^2)$$ Hence, since $\displaystyle\lim_{x\to 0}u(x)=0$, we see than an $\underset{u\to 0}o(u^2)$ function is an $o(x^2)$ when $x$ tends to $0$.

This leads to $$\begin{align*}(1+x)^{1/x} & = e\left(1+\left(-\dfrac x2+\dfrac{x^2}3+\underset{x\to 0}{o}(x^2)\right)+\dfrac 12\left(\dfrac{x^2}4+\underset{x\to 0}o(x^2)\right)+\underset{x\to 0}{o}(x^2)\right)\\ & = e-\dfrac{ex}2+\dfrac{11e}{24}x^2+\underset{x\to 0}o(x^2)\end{align*}$$ And so : $$\dfrac{(1+x)^{1/x}-e+\dfrac{ex}2}{x^2}=\dfrac{11e}{24}+\underset{x\to 0}o(1)$$ which means exactly that $$\lim_{x\to 0}\dfrac{(1+x)^{1/x}-e+\dfrac{ex}2}{x^2}=\dfrac{11e}{24}$$

Philippe Malot
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