Is it possible to prove simply by manipulation that $\lim\limits_{x\to\infty} x^2\bigl(2-e^{-i/x}-e^{i/x}\bigr)=1$?

Question

In fact, it turns out that $$\lim_{x\to\infty} x^2\left(2-e^{-z/x}-e^{z/x}\right)=-z^2$$ for any complex $z$. What is a simple (no DLH, Taylor, Maclaurin, ...) and "primitive" (no using directly or indirectly notable trigonometric limits, namely $\lim\limits_{x\to0}\frac{\sin x}{x}=\lim\limits_{x\to0}\frac{\tan x}{x}=1$ and $\lim\limits_{x\to0}\frac{1-\cos^z x}{x}=z/2$) approach to this problem? I'm asking out of curiosity, after doing some homework related to, indeed, notable limits, and derivatives. Unfortunately I couldn't do much, thanks for any help.

Answer 1

It holds that $$ x^2(2-e^{-z/x}-e^{z/x}) = -x^2(e^{z/(2x)}-e^{-z/(2x)})^2=-z^2\Bigl(\frac{\sinh(z/(2x))}{z/2x}\Bigr)^2. $$ Are you allowed to use that $\frac{\sinh t}{t}\to 1$ as $t\to 0$, then this is finished... If not, then I must ask you: What are you allowed to use?

Answer 2

$(2-e^{-i/x}-e^{i/x})=2-(2 cos{\frac{1}{x}})$

but $cos{\frac{1}{x}}=\dfrac{1-(\tan{\frac{1}{2x}})^2}{1+(\tan{\frac{1}{2x}})^2}$

Then you have

$(2-e^{-i/x}-e^{i/x})=\dfrac{4(\tan{\frac{1}{2x}})^2}{1+(\tan{\frac{1}{2x}})^2}$

Can we finish here with your limitations?

Answer 3

$$ \lim_{x\to\infty} x^2\left(2-e^{-z/x}-e^{z/x}\right)= \lim_{x\to\infty} x^2\left[-\left(e^{z/2x}-e^{-z/2x}\right)^2\right] \\ = \lim_{x\to\infty} -x^2\left(2\sinh\left(\frac{z}{2x}\right)\right)^2 \\ = \lim_{x\to\infty} -\left(2x\sinh\left(\frac{z}{2x}\right)\right)^2$$ I'm not sure what your limitations are from here on out, but if you can now show that $$\lim_{x\to\infty}2x\sinh\left(\frac{z}{2x}\right)=z$$ then it'll follow pretty easily.