#### Answer

The solution is
$$\lim_{x\to\infty}\csc\frac{1}{x}(e^{1/x}-1)=1.$$

#### Work Step by Step

To solve this limit we will use L'Hopital's rule. "LR" will denote "Apply L'Hopital's rule".
$$\lim_{x\to\infty}\csc\frac{1}{x}(e^{1/x}-1)=\lim_{x\to\infty}\frac{e^{\frac{1}{x}}-1}{\sin\frac{1}{x}}=\left[\frac{e^{\frac{1}{\infty}}-1}{\sin\frac{1}{\infty}}\right]=\left[\frac{e^0-1}{\sin0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\infty}\frac{\left(e^{\frac{1}{x}}-1\right)'}{\left(\sin\frac{1}{x}\right)'}=\lim_{x\to\infty}\frac{e^{\frac{1}{x}}\left(\frac{1}{x}\right)'}{\cos\frac{1}{x}\left(\frac{1}{x}\right)'}=\lim_{x\to\infty}\frac{e^{\frac{1}{x}}}{\cos\frac{1}{x}}=\left[\frac{e^{1/\infty}}{\cos(1/\infty)}\right]=\frac{e^0}{\cos0}=1.$$