Question

The problem is attached

Answer 1

the answer is d.

Answer 2

Note that for large values of \(x\) the value of \(e^{5/2x}\) gets arbitrarily small so we can use a taylor expansion to simplify this expression. Note that:
\[\Large e^x = 1 + x + \frac{x^2}{ 2}+ \dots \] such that: \[\Large e^{\frac{5}{2x}}\sim 1+ \frac{5}{2x} \]
We drop the higher order approximations because they vanish fast to zero, plugging this back in our equation gives:
\[\Large \lim_{x \to \infty} xe^{\frac{5}{2x}}-x=x \left( 1 + \frac{5}{2x}\right)-x = \lim_{x \to \infty} \frac{5}{2}=\frac{5}{2} \]

Answer 3

Those problems are very cool Professor @eliassaab, keep them coming.

Answer 4

well notice$$x(e^{5/2x}-1)=x\left(\frac5{2x}+\frac12\left(\frac5{2x}\right)^2+\dots+\frac1{n!}\left(\frac5{2x}\right)^n+\dots\right)=\frac52+x\cdot(\dots)$$

Answer 5

Here is the answer generated by http://saab.org

Answer 6

@Spacelimbus , I am glad your enjoying them. I posted another one.