Question

$\underset{x\to \infty }{lim}{\{{\left(\frac{x}{x+1}\right)}^a+\sin \frac{1}{x}\}}^a$ is equal to

• A. $e^{a-1}$
• B. $e^{1-a}$
• C. $e$
• D. $0$

Answer 1

The correct option is A $e^{a-1}$

${lim}_{x\to \infty }{\left\{{\left(\frac{x}{x+1}\right)}^a+\sin \frac{1}{x}\right\}}^a$

$\Rightarrow {lim}_{x\to \infty }{e}^{a\log \left({\left(\frac{x}{x+1}\right)}^a+\sin \frac{1}{x}\right)}$

$\Rightarrow {lim}_{x\to \infty }{e}^{a\log \left(\frac{1}{1+\frac{a}{x}+.......}+\frac{\left(\frac{1}{x}+\frac{1}{3!}\frac{1}{x^3}+.......\right)}{1}\right)}$

$\Rightarrow {lim}_{x\to \infty }{e}^{a\log \left(1+\left(1+\frac{a}{x}+.......\right)\left(\frac{1}{x}+\frac{1}{3!}\frac{1}{x^3}+........\right)\right)-\log \left(1+\frac{a}{x}+......\right)}=e^{a-1}$