Attempt to solve $\lim_{x\to +\infty} \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}$ I tried to solve this limit. Is it correct? There exists a more straightforward way?
$$\lim_{x\to +\infty} \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}$$
$$\lim_{x\to +\infty} \exp{\left(\frac{x^2-1+1}{1+x}\right)} -\exp\left(x^2\ln\left(1+\frac1x\right)\right)$$
$$\lim_{x\to +\infty} \exp(x-1)\cdot\exp{\left(\frac{1}{1+x}\right)} -\exp\left(x^2\left(\frac1x-\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)\right)\right)$$
$$\lim_{x\to +\infty} \exp(x-1)\cdot{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)} -\exp\left(x-\frac{1}{2}+o(1)\right)$$
$$\lim_{x\to +\infty} \exp\left(-\frac12\right)\cdot\exp\left(x-\frac12\right)\cdot{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)} -\exp\left(x-\frac{1}{2}+o(1)\right)$$
$$\lim_{x\to +\infty}\underbrace{\exp\left(x-\frac12\right)}_{\to+\infty}\left[\exp\left(-\frac12\right)\cdot{\underbrace{\left(1+\frac{1}{1+x}+o\left(\frac{1}{x}\right)\right)}_{\to0}} -\underbrace{\exp\left(o\left(1\right)\right)}_{\to 1}\right] $$
$$= +\infty\cdot\left[0-1\right] = -\infty$$
 A: You are correct. A slightly shorter way with a more general result: let $a$ be a real number then, as $x\to +\infty$,
$$\begin{align}
\exp{\left(\frac{x^2}{a+x}\right)}& - \left(1+\frac1x\right)^{x^2}=
\exp{\left(\frac{x}{1+\frac{a}{x}}\right)}-\exp{\left(x^2\ln\left(1+\frac{1}{x}\right)\right)}\\
&=\exp{\left(x\left(1-\frac{a}{x}+o\Big(\frac{1}{x}\Big)\right)\right)}-\exp{\left(x^2\left(\frac{1}{x}-\frac{1}{2x^2}+o\Big(\frac{1}{x^2}\Big)\right)\right)}\\
&=e^x\left(e^{-a+o(1)}-e^{-1/2+o(1)}\right)\to 
\begin{cases}
+\infty & \text{if $a<1/2$}\\
-\infty & \text{if $a>1/2$}
\end{cases}
\end{align}
$$
For the case $a=1/2$, we need a more precise expansion and the result is $-\infty$.
A: I always try to find a solution without using $o(f(x))$ (yes, I know that De L'Hopital is equivalent to $o(f(x))$)
\begin{align*}
& \lim_{x\to +\infty}\left[ \exp{\left(\frac{x^2}{1+x}\right)} - \left(1+\frac1x\right)^{x^2}\right] = \\
& \lim_{x\to +\infty}\left[ \exp{\left(\frac{x^2}{1+x}\right)} - \exp\left(x^2\log\left(1+\frac1x\right)\right)\right] = \\
& \lim_{y\to 0^+}\left[ \exp{\left(\frac{1}{y+y^2}\right)} - \exp\left(\frac{\log\left(1+y\right)}{y^2}\right)\right] = \\
& \lim_{y\to 0^+}\left[ \exp{\left(\frac{1}{y}-\frac{1}{1+y}\right)} - \exp\left(\frac{1}{y}+\frac{\log\left(1+y\right)-y}{y^2}\right)\right] = \\
& \lim_{y\to 0^+} \exp\left(\frac{1}{y}\right) \left[ \exp{\left(-\frac{1}{1+y}\right)} - \exp\left(\frac{\log\left(1+y\right)-y}{y^2}\right)\right] = (*)
\end{align*}
and given that
$$
\lim_{y\to0}\frac{\log\left(1+y\right)-y}{y^2}=\lim_{y\to0}\frac{\frac{1}{1+y}-1}{2y}=-\frac{1}{2}
$$
we have
$$
(*) = \lim_{y\to0^+}\exp\left(\frac{1}{y}\right)[e^{-1}-e^{-1/2}] = -\infty
$$
