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I came across this problem a few days ago and I have not been able to solve it. Wolfram Alpha says the answer is 1/2 but the answer I came up with is 0. Can anyone see what is wrong with my work and/or provide the correct way of solving this problem?

$$\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x} $$ $$\lim_{x \to \infty} \frac{ex^{x+1}-x[(x)(1+\frac{1}{x})]^x}{[(x)(1+\frac{1}{x})]^x} $$ $$\lim_{x \to \infty} \frac{ex^{x+1}-x^{x+1}(1+\frac{1}{x})^x}{x^x(1+\frac{1}{x})^x} $$ $$\lim_{x \to \infty} \frac{ex^{x+1}-x^{x+1}e}{x^xe} $$ $$\lim_{x \to \infty} \frac{x-x}{1} $$ $$\lim_{x \to \infty} \frac{0}{1} $$ $$0$$

I understand my mistakes may be simple and trivial, but I'm trying to learn. Thank you for your help!

Shuri2060
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Brock Boehler
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    In the fourth line you are "partially" taking a limit, by replacing a Standard Limit by $e$, and then you continue to calculate the rest of the limit. This is not right. You can only calculate "part of a limit" if you were, for example, able to write the given limit as a product of two limits, one of them being $(1+1/x)^x$, but you can't do that here. – imranfat Aug 20 '17 at 17:51
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    @imranfat I think that should be an answer. To give a simpler example: $\lim_{x\to0}\frac{x}{x}\neq\lim_{x\to0}\frac{0}{x}$ – Shuri2060 Aug 20 '17 at 17:59

4 Answers4

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Change $x=\frac{1}{t}$ and simplify: $$L=\lim_{t\to 0} \frac{e-(1+t)^{1/t}}{t(1+t)^{1/t}}=L'H=$$ $$\lim_{t\to 0} \frac{(1+t)\ln{(1+t)}-t}{t^2(1+t)+t^2-t(1+t)\ln{(1+t)}}=Taylor=$$ $$\lim_{t\to 0} \frac{\frac{t^2}{2}+O(t^3)}{t^2+O(t^3)}=\frac{1}{2}.$$

farruhota
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  • It is rare to find usage of both the tools (Taylor and L'Hospital's Rule) together. Normally one prefers one tool over another. – Paramanand Singh Aug 20 '17 at 21:55
  • @Paramanand Singh, thank you for your comment. One should not restrict himself and use a convenient tool. In the second step I preferred one Taylor over two L'Hs. – farruhota Aug 21 '17 at 05:02
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This solution is not as elegant as @farruhotas, but it's more direct. This solution relies on Taylor approximations and handwaving. I realize it's not rigorous; the purpose of this is providing intuition as to why the limit is $\frac{1}{2}$. (I'll be assuming that $x$ is an integer. Assuming the limit exists, since $x \to \infty$, this isn't an issue.)

Consider the numerator. The highest power terms are $x^{x+1}$ terms. We will first show that the $x^{x+1}$ terms cancel as $x \to \infty$. Consider the expansion of $(x+1)^x$:

$$ (x+1)^x = x^x + \binom{x}{1} x^{x-1} + \binom{x}{2} x^{x-2} + \cdots \binom{x}{\lfloor x/2 \rfloor} x^{x-\lfloor \frac{x}{2} \rfloor}+ \cdots + \binom{x}{x-1} x^1 + 1 $$

Its highest power terms are of order $x^x$. Since the $x^k$ term of $\binom{x}{k}$ is approximately $\frac{1}{k!} x^x$, the $x^x$ term of $(x+1)^x$ is approximately

$$ x^x (1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{x!} ) \approx x^x e$$

as $x \to \infty$. Thus, the $x^{x+1}$ term of $x(x+1)^x$ is $ex^{x+1}$. So the $x^{x+1}$ powers in the numerator cancel.

Now we need to find the $x^x$ powers in the numerator. The first summand $ex^{x+1}$ doesn't provide any, so let's look at the second summand $x(x+1)^x$. Go back to the expansion of $(x+1)^x$:

$$ (x+1)^x = x^x + \binom{x}{1} x^{x-1} + \binom{x}{2} x^{x-2} + \cdots \binom{x}{\lfloor x/2 \rfloor} x^{x-\lfloor \frac{x}{2} \rfloor}+ \cdots + \binom{x}{x-1} x^1 + 1 $$

Now we are looking for all terms with $x^{x-1}$ powers. Since the $x^k$ term of $\binom{x}{k}$ is approximately

$$ \frac{-1-2-\cdots -k}{k!} = -\frac{k(k-1)}{2 k!} = -\frac{1}{2} \frac{1}{(k-2)!}$$

we have that the $x^{x-1}$ term of $(x+1)^x$ is approximately

$$ x^{x-1} \cdot (-\frac{1}{2}) (0 + 0 + 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots) \approx x^{x-1} (-\frac{e}{2}) $$

as $x \to \infty$. Thus, the $x^{x}$ term in the numerator is

$$ 0 + x \cdot (x^{x-1} (-\frac{e}{2} )) = \boxed{\frac{e}{2} x^{x}} .$$

Now consider the denominator. We already know that the $x^x$ term of $(x+1)^x$ tends to $e x^x$ as $x \to \infty$.

So, the limit becomes

\begin{align*} \lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x} &= \lim_{x \to \infty} \frac{\frac{e}{2} x^x }{e x^x} \\ &= \frac{1}{2}. \end{align*}

No elegant tricks. Just Taylor expansions and handwaving.

aras
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The expression can be written as $$x\left(e\left(1+\frac{1}{x}\right)^{-x}-1\right)$$ and then we put $x=1/t$ to get $$\frac{e(1+t)^{-1/t}-1}{t}$$ This can be further expressed as $$\frac{e^{u} - 1}{u}\cdot\frac{u}{t}$$ where $u=1-\dfrac{\log(1+t)}{t}\to 0$ and thus $(e^{u} - 1)/u\to 1$. It follows that the desired limit is $$\lim_{t\to 0^{+}}\frac{t-\log(1+t)}{t^{2}}$$ which is easily evaluated as $1/2$ via Taylor series or L'Hospital's Rule.

The mistake in your solution is very common. In general one can not replace a sub-expression by its limit while evaluating the limit of an expression. Thus your replacement of $(1+(1/x))^{x}$ with $e$ is wrong. Such replacements are allowed only in two specific scenarios.

Paramanand Singh
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Consider $$y=\frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}=\frac{ex^{x+1}}{(x+1)^x}-x$$ Now $$z=\frac{ex^{x+1}}{(x+1)^x}\implies \log(z)=1+(x+1)\log(x)-x\log(x+1)=1+\log(x)-x \log\left(1+\frac 1x\right)$$ Now, using Taylor $$\log\left(1+\frac 1x\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$ makes $$\log(z)=\log \left({x}\right)+\frac{1}{2 x}-\frac{1}{3 x^2}+O\left(\frac{1}{x^3}\right)$$ $$z=e^{\log(z)}=x+\frac{1}{2}-\frac{5}{24 x}+O\left(\frac{1}{x^2}\right)$$ which finally makes $$y=\frac{1}{2}-\frac{5}{24 x}+O\left(\frac{1}{x^2}\right)$$ showing the limit and how it is approached.

Using $x=10$ which is really small, the exact value would be $$\frac{100000000000 e}{25937424601}-10\approx 0.480153$$ while the above approximation would simply give $\frac{23}{48}\approx 0.479167$.

Claude Leibovici
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