$$\lim\limits_{x\to \infty}{\left( \frac{(1+x)^x}{x^xe}\right)}^x$$$$=\lim\limits_{x\to \infty}{\left( 1+\frac1x\right)}^{x^2}e^{-x}$$ Now since $\lim\limits_{x\to \infty}{\left( 1+\frac1x\right)}^{x}$ is $e$ when $x \to \infty$, and that raised to the power of $x$ is $e^x$ $$=\lim\limits_{x\to \infty}e^xe^{-x}=1$$
However, the answer is given $\frac {1}{\sqrt e}$. How? And where am I going wrong?
On
Another way of evaluating the limit is by taking the natural log, we let $$L= \lim_{x\rightarrow \infty} \left(\frac{(1+x)^x}{x^xe}\right)^{x},$$ then \begin{align*} \ln L &= \lim_{x\rightarrow \infty} \left(x^2\ln \left(1+\frac{1}{x}\right)-x\right) \\ &=\lim_{x\rightarrow \infty} \frac{\ln \left(1+\frac{1}{x}\right)-\frac{1}{x}}{\frac{1}{x^2}} \\ &\overset{\textrm{L'H}}{=} \lim_{x\rightarrow \infty} \frac{\frac{-\frac{1}{x^2}}{1+\frac{1}{x}}+\frac{1}{x^2}}{-\frac{2}{x^3}} \\ &= \lim_{x\rightarrow \infty} \frac{-\frac{x^2}{1+x}+x}{-2} \\ &= \lim_{x\rightarrow \infty} -\frac{1}{2} \cdot \left(1-\frac{1}{x+1}\right) \\ &= -\frac{1}{2}. \end{align*} Hence, $L=\boxed{\frac{1}{\sqrt{e}}} . $
As pointed out in the comments, you cannot do this due to the limit involving the variable as a power itself. However, if you insist you can use the $1^\infty$ form which you have used here with a little bit of manipulation. Write the function as
$$\lim_{x \to \infty}\left( \frac{1+\frac1x}{e^{\frac1x}}\right)^{x^2}$$ which is in the form of $1^\infty$.
Now the limit is $$\lim_{x \to \infty}e^{x^2\frac{\left(1+\frac1x-e^{\frac1x}\right)}{e^{\frac1x}}}$$$$=\lim_{x \to \infty}e^{x^2+x-x^2e^\frac1x}$$ Let $y=\frac1x$.$$\lim_{y \to 0}e^{\frac{1}{y^2}+\frac1y-\frac{e^y}{y^2}}$$ Evaluating the power using Taylor Expansion, $$\lim_{y \to 0}\left(\frac{1}{y^2}+\frac1y-\frac{e^y}{y^2} \right)$$$$=\lim_{y \to 0}\left(\frac{1}{y^2}+\frac1y-\frac1{y^2}-\frac 1y -\frac12 -\frac y{3!}-\frac {y^2}{4!}- \dots \right)$$$$=-\frac12$$
So the answer is $e^{-\frac12}=\frac1{\sqrt e}$.