Question

Do the problem in the attached file

Answer 1

\[=\lim_{x \rightarrow \infty}x(e^{\frac{5}{2x}}-1)=\lim_{x \rightarrow \infty}\frac{e^{\frac{5}{2x}}-1}{\frac{1}{x}}\]Now it is in a form in which you can use l'Hospital's rule. After apply it once and simplifying, I get 5/2 as the limit.

Answer 2

5/2 is the answer.