real analysis - Finding limit from Squeeze theorem.

I recently came across a problem which stated to find the limit of an equation through the Squeeze theorem,

$$\lim_{n\to\infty} \left(\frac{2n-5}{3n+1}\right) ^{n}$$
My approach: I did the question with L'Hospital's Rule just for the sake of finding the limit,

$$\log (L) = n(\log(2n-5) - \log(3n+1))$$
$$\log(L) = \frac{\log(2n-5) - \log(3n+1)}{\frac{1}{n}}$$
By differentiating,

$$\log(L) = \frac{\frac{2}{2n-5}-\frac{3}{3n+1}}{\frac{-1}{n^2}}$$
$$\log(L) = -\frac{17}{12}$$
$$L = e^{-\frac{17}{12}}$$
This was the limit obtained by me. But I wasn't able to approach through Squeeze Theorem.

Answer

$$0<\left(\frac{2n-5}{3n+1}\right) ^{n}<\left(\frac{2}{3}\right)^n,$$
so
$$\lim_{n\to\infty} \left(\frac{2n-5}{3n+1}\right) ^{n}=0.$$