Why is it wrong to solve the limit this way? Consider the following limit：
\begin{align*}
&\lim_{x\to +\infty}\frac{e^x}{\left(1+\frac1x\right)^{x^2}}\\
&=e^{\lim_{x\to +\infty}(x-x^2\ln\left(1+\frac1x\right))}\\
&=e^{\lim_{x\to +\infty}x^2(\frac1x-\ln\left(1+\frac1x\right))}\\
&=e^{\lim_{x\to 0^+ }\frac1{x^2}(x-\ln(1+x))}\\
&=e^{\lim_{x\to 0^+ }\frac{\frac12x^2}{x^2}}\\
&=e^{\frac12}
\end{align*}
The above solution is correct and another solution is following:
\begin{align*}
&\lim_{x\to +\infty}\frac{e^x}{(1+\frac1x)^{x^2}}\\
&=\lim_{x\to +\infty}\frac{e^x}{[(1+\frac1x)^{x}]^x}\\
&=\lim_{x\to +\infty}\frac{e^x}{e^x}\\
&=1
\end{align*}
Why is that wrong? Is that because we didn't take the limit of both the numerator and the denominator?
 A: It is not allowed, in general, take the limit for a single part of the entire expression.
Notably, in this case you have made this mistake twice in two different ways using that
$$\lim_{x\to +\infty}\frac{f(x)}{g(x)} =\lim_{x\to +\infty}\frac{f(x)}{\lim_{x\to +\infty}g(h(x))}$$
and then
$$\lim_{x\to +\infty}g(h(x)) =g(\lim_{x\to +\infty}h(x))$$
For a general discussion refer to:

*

*Analyzing limits problem Calculus (tell me where I'm wrong).
and other similar examples.

For the solution we can avoid Taylor's series or l'Hospital as follows
$$\frac{e^x}{\left(1+\frac1x\right)^{x^2}} = e^{-\frac{\log\left(1+\frac1x\right)-x}{\frac1{x^2}}} \to e^{-\frac12}$$
using the method presented here.
A: (I've colored equals signs below as red when the argument is falty. Apologies to the color-blind.)
You've done this:
$$\lim_{x\to\infty} \frac{f(x)}{g(x)^x}\color{red}=\lim_{x\to\infty}\frac{f(x)}{(\lim_{y\to\infty} g(y))^x}$$
If you could do that kind of substitution, you could solve a lot of problems this way:
$$\lim_{x\to \infty}\left(1+\frac1x\right)^x\color{red}=\lim_{x\to\infty}\left(\lim_{y\to\infty} 1+\frac 1y\right)^x=\lim_x 1^x=1.$$

The real question is why you'd think you could let one tiny part of the limit go to infinity first.
One way to see this is wrong is to compute:
$$\lim_{x\to\infty}\frac{g(x)^x}{(\lim_{y\to\infty} g(y))^x}$$
In your technique, this limit is $1.$
If $L=\lim_{y\to\infty} g(y),$ then let $h(x)=g(x)-L.$ $h(x)$ is the "error."
Since $g(x)=L+h(x),$ we can compute the limit:
$$\lim_{x\to\infty} \frac{g(x)^x}{L^x}=\lim\left(\frac{g(x)}{L}\right)^x=\lim\left (1+\frac{h(x)}{L}\right)^x.$$
All we really know is that $h(x)\to 0.$ So, for example, if $h(x)=\frac 1x,$ then the right side of the limit would be $e^{1/L}.$
In your case, you need an estimate for $h(x)=e-(1+1/x)^x.$ I'm not sure how to do that, but I'd bet, from the actual answer, that $h(x)=\frac{e}{2x}+o\left(\frac1x\right).$

Another way to see the error is to take the first approach, but apply the second answer's logic. We'll just take the logarithms. Your second logic is:
$$\lim_{x\to\infty} (x-x^2\log(1+1/x))\color{red}=\lim_{x\to\infty} x-x\left(\lim_{y\to\infty} y\log(1+1/y)\right)=0.$$
since $y\log(1+1/y)\to 1.$
As you can see here, the common factor of $x$ is the problem. It is true that $1-x\log(1+1/x)\to 0,$ but it is not true that $x(1-x\log(1+1/x))\to 0.$
