Finding the limit of $\frac{e^{x}+x-\cos(2x)}{x^2}$ How would one find the limit for the following problem.
$x\rightarrow\infty$
$\frac{e^{x}+x-\cos(2x)}{x^2}$
I did the hospital rule.
$\frac{e^x+1+2\sin(2x)}{2x}$
But now I am stuck I did this but I feel it diverges.
$e^x+1+2\sin(2x)*\frac{1}{2x}$

Since $\cos(2x) \leq 1,\,\forall x$, and $e^x = 1+x+x^2/2+x^3/6+\cdots \geq 1+x+x^2/2+x^3/6$, we obtain, $$\frac{e^{x}+x-\cos(2x)}{x^2} \geq \frac{2x+x^2/2+x^3/6}{x^2} \geq \frac{x^3/6}{x^2} \geq \frac{x}{6},\,\forall x > 0.$$ The lower bound diverges, hence the original series.
