<s>[INST] Can't understand why cosh(1/x) changes the result for this limit I have this limit:
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh{\frac 1x}\over x^2}-1\right)$$
If I solve this limit by hand, I get 1
$\require{cancel}$
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh{\cancelto{0}{\frac 1x}}\over x^2}-1\right)=$$
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cancelto{1}{\cosh{0}}\over x^2}-1\right)=$$
$$\lim_{x \to -\infty} \cancel{x^2}\left({\cancel{x^2}+1\cancel{-x^2}\over \cancel{x^2}}\right)=1$$
However, both Wolfram Alpha and the exercise say the correct result is ${\frac 32}$
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh{\frac 1x}\over x^2}-1\right) = \frac 32$$
However, if I remove $\cosh{\frac 1x}$, which for ${x \to -\infty}$ tends to $1$, suddenly Wolfram Alpha says the result is 1:
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\over x^2}-1\right) = 1$$
I'm at a complete loss. Wolfram Alpha uses L'Hopital's rule to solve the first limit, so it doesn't help me undersand the discrepancy
 [/INST] Note that the Taylor series of $\cosh(z)$ at $z=0$  is $1+\frac{z^2}{2}+o(z^2)$. Hence, as $x\to -\infty$, we find
$$\begin{align}
x^2\left({(x^2+1)\cosh\left({\frac 1x}\right)\over x^2}-1\right)
&=x^2\left({(x^2+1)(1+\frac{1}{2x^2}+o(1/x^2))\over x^2}-1\right)
\\
&=(x^2+1)\left(1+\frac{1}{2x^2}+o(1/x^2)\right)-x^2\\
&=1+\frac{1}{2}+o(1)\to \frac{3}{2}.\end{align}$$
In your work you implicitly used a less precise Taylor series, namely $1+o(1)$, which yields
$$x^2\left({(x^2+1)(1+o(1))\over x^2}-1\right)=(x^2+1)(1+o(1))-x^2=1+x^2\cdot o(1)+o(1)$$
and we can't conclude that the limit is $1$ because $x^2\cdot o(1)$ is an indeterminate form $+\infty\cdot 0$.
</s>