Why Using Maclaurin series is giving me a different answer?

Question

$$\lim_{x \to\infty }\frac{(1+\frac{1}{x})^{x^{2}}}{e^{x}}= \frac{1}{\sqrt{e}}$$ I have to proove this equation using Maclaurin expansion, which I know how to do. However, my question is when looking at the numerator.It's equal to $e^{x}$ then divided by denominator is equal to 1, meaning RHS is not equal to LHS. My question is more of a technical then how to solve this,because I know how. Using both ways I get different answers for some reason.

****Edit*** $(1+\frac{1}{x})^{x}=e$

and then outer bracelet: $e^x$

following the answers I got a little bit confused... why $\lim_{x \to \infty } ((1+\frac{1}{x})^{x})^{constant}=e^{constant}$ but does not hold true for $\lim_{x \to \infty } ((1+\frac{1}{x})^{x})^{x}=e^{x} ?$

Answer 1

The Mclaurin series for $(1+t)^{1/t^2}$ is $e^{1/t-1/2+t/3}$, let $x=1/t$, then your limit becomes

$$L=\lim_{t \rightarrow 0} \frac{(1+t)^{1/t^2}}{e^{1/t}}= \lim_{t \rightarrow 0} \frac{e^{1/t-1/2+t/3}}{e^{1/t}}=e^{-1/2}.$$

Answer 2

You are (probably) using

$$\lim_{x\to\infty}\frac{\left(1+\frac1x\right)^{x^2}}{e^x}=\lim_{x\to\infty}\frac{\left(\lim_{y\to\infty}\left(1+\frac1y\right)^y\right)^x}{e^x}.$$

This is not justified.

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In blue, $\dfrac{(1+\frac1x)^{x^2}}{e^x}$ and in green $\dfrac{e^x}{e^x}$.