How would one evaluate this limit without using L'Hôpital's rule? What I know for sure is that this limit equals to zero, but I don’t know how to solve it.
$$ \lim_{x\rightarrow \infty}\left(\frac{2x-5}{2x-2}\right)^{4x^{2}} = 0 $$
Edit: This is Wolframalpha solution: https://www.wolframcloud.com/objects/f6d89370-5d6b-44de-9f4e-0377dea18aad , but it uses the L'Hôpital's rule and I think it is a quite complicated solution.
$\lim_\limits {n\to \infty} (1+\frac {x}{n})^n = e^x$
$\lim_\limits {x\to \infty} (\frac {2x-5}{2x-2})^{4x^2}\\ \lim_\limits {x\to \infty} (1 +\frac{-3}{2x-2})^{4x^2}$
Now $\lim_\limits {x\to \infty} (1 +\frac{-3}{2x-2})^{2x-2} = e^{-3}$
So, what do you have to do to get a factor like this out of the line above?
$\lim_\limits {x\to \infty} ((1 +\frac{-3}{2x-2})^{2x-2})^{2x+2}(1 +\frac{-3}{2x-2})^4\\ \lim_\limits {x\to \infty} (e^{-3})^{2x+2}(1 +\frac{-3}{2x-2})^4$
$0<e^{-3} < 1$
What I have done here is not rigorous.
But you can use the work above to say:
$0\le \lim_\limits {x\to \infty}(\frac {2x-5}{2x-2})^{4x^2} \le \lim_\limits {x\to \infty} e^{-6x}$
$\lim_\limits {x\to \infty}(\frac {2x-5}{2x-2})^{4x^2} = 0$ by the squeeze theorem.