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
 A: 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 $+\infty\cdot 0$.
A: Your step is not allowed since, in general, we can't take the limit for a single part of the entire expression. For this important issue refer to the related:

*

*Analyzing limits problem Calculus (tell me where I'm wrong).
As an alternative to Taylor's expansion, we can solve the limit as follows using that

*

*$\cosh t = \frac{e^t + e^{-t}}{2}$

*$\frac{e^t-t-1}{t^2} \to \frac12$ as $t \to 0$ (ref. here)

to obtain
$$x^2\left({(x^2+1)\cosh{\frac 1x}\over x^2}-1\right)= (x^2+1)\frac{e^{\frac1x} + e^{-\frac1x}}{2}-x^2=$$
$$=\frac{e^{\frac1x} + e^{-\frac1x}}{2}+x^2\left(\frac{e^{\frac1x} + e^{-\frac1x}}{2}-1\right)\to 1+\frac12 =\frac32$$
indeed $\frac{e^{\frac1x} + e^{-\frac1x}}{2} \to 1$ and
$$x^2\left(\frac{e^{\frac1x} + e^{-\frac1x}}{2}-1\right) =x^2\left(\frac{e^{\frac1x} -\frac1x -1+ e^{-\frac1x}+\frac1x-1}{2}\right)=$$
$$=\frac12\left(\frac{e^{\frac1x} -\frac1x -1}{\frac1{x^2}}+\frac{e^{-\frac1x} +\frac1x -1}{\frac1{x^2}}\right) \to \frac12 \left(\frac12 + \frac 12\right)= \frac12$$
A: Since we are always closer to $0$ than to $\infty$, it would have been much simpler to use
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh\left(\frac 1x\right)\over x^2}-1\right)=\lim_{t \to 0^-}\frac{\left(t^2+1\right) \cosh (t)-1}{t^2}$$ Using Taylor series around $t=0$
$$\frac{\left(t^2+1\right) \cosh (t)-1}{t^2}=\frac{\left(t^2+1\right) \Big[1+\frac{t^2}{2}+\frac{t^4}{24}+O\left(t^6\right) \Big]-1}{t^2}$$
$$\frac{\left(t^2+1\right) \cosh (t)-1}{t^2}=\frac{\frac{3 t^2}{2}+\frac{13 t^4}{24}+O\left(t^6\right) }{t^2}=\frac{3}{2}+\frac{13 t^2}{24}+O\left(t^4\right)$$
