Evaluating $\lim_{x\to \infty}\Biggl(\dfrac{(x+1)^x}{x^x\cdot\mathcal{e}}\Biggr)^x$

Question

I'm stuck in this problem where my answer is repeatedly resulting in1, which isn't correct, which is making me doubt my procedure. Please point out where I went wrong. $$\lim_{x\to \infty}\Biggl(\dfrac{(x+1)^x}{x^x\cdot\mathcal{e}}\Biggr)^x=\lim_{x\to \infty}\dfrac{(1+\frac{1}{x})^{x^2}}{\mathcal{e}^x}=\lim_{x\to \infty}\dfrac{\mathcal{e}^{\frac{1}{x}\cdot x^2}}{\mathcal{e}^x}\text{$\Bigl(\because1^{\infty}$ form}\Bigr)$$ $$\lim_{x\to \infty}\dfrac{\mathcal{e}^{\frac{1}{x}\cdot x^2}}{\mathcal{e}^x}=1$$ That's my answer, but it is supposed to be $\dfrac{1}{\sqrt{\mathcal{e}}}$.

Answer 1

Here is an explanation why the OP was wrong.

Since$$(1+\frac{1}{x})= \mathcal{e}^{\frac{1}{x}}$$ is not true, therefore we can't be sure whether$$\lim_{x\to \infty}\dfrac{(1+\frac{1}{x})^{x^2}}{\mathcal{e}^x}=\lim_{x\to \infty}\dfrac{\mathcal{e}^{\frac{1}{x}\cdot x^2}}{\mathcal{e}^x}$$ is true or not.

---

Note:

If we want to use the theorem

$$\lim_{x\to a}f(x)^{g(x)}=\mathcal{e}^{\lim_{x\to a}(f(x)-1)g(x)} \;\;\text{ if }\; f(a) \to 1 \;\;\text{ and } \;g(a) \to \infty$$

we have to write$$\lim_{x\to\infty}\dfrac{(1+\frac{1}{x})^{x^2}}{\mathcal{e}^x}=\lim_{x\to\infty}\left(\dfrac{1+\frac{1}{x}}{\mathcal{e}^{1/x}}\right)^{x^2}$$ instead of $$\lim_{x\to \infty}\dfrac{(1+\frac{1}{x})^{x^2}}{\mathcal{e}^x}=\lim_{x\to \infty}\dfrac{\mathcal{e}^{\frac{1}{x}\cdot x^2}}{\mathcal{e}^x}.$$

Answer 2

The logarithm of your expression is $$x^2\ln\left(1+\frac1x\right)-x =x^2\left(\frac1x-\frac1{2x^2}+O(x^{-3})\right)-x =-\frac12+O(x^{-1})\to-\frac12$$ as $x\to\infty$.

Answer 3

A little more generally, let $f(a) =\lim_{x\to \infty}\Biggl(\dfrac{(x+a)^x}{x^x\cdot\mathcal{e}}\Biggr)^x $. Then

$\begin{array}\\ g(a) &=\ln(f(a))\\ &=\lim_{x\to \infty}x\ln\dfrac{(x+a)^x}{x^x\cdot\mathcal{e}}\\ &=\lim_{x\to \infty}x\left(\ln\dfrac{(x+a)^x}{x^x}-1\right)\\ &=\lim_{x\to \infty}\left(x\ln\dfrac{(x+a)^x}{x^x}-x\right)\\ &=\lim_{x\to \infty}\left(x^2\ln\dfrac{(x+a)}{x}-x\right)\\ &=\lim_{x\to \infty}\left(x^2\ln(1+\frac{a}{x})-x\right)\\ &=\lim_{x\to \infty}\left(x^2(\frac{a}{x}-\frac{a^2}{2x^2}+O(\frac1{x^3}))-x\right)\\ &=\lim_{x\to \infty}\left(ax-\frac{a^2}{2}+O(\frac1{x})-x\right)\\ &=\lim_{x\to \infty}\left((a-1)x-\frac{a^2}{2}+O(\frac1{x})\right)\\ \end{array} $

If $a=1$ (as in this problem), $g(1) =-\frac12$.

If $a \ne 1$, then $\lim_{x \to \infty} (g(a)-(a-1)x) =-\frac{a^2}{2} $.