Question

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

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

Answer 1

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

Let $y=\frac{1}{x}$
$\underset{x\to \infty }{lim}{\{{\left(\frac{x}{x+1}\right)}^a+\sin \frac{1}{x}\}}^x$

$=\underset{x\to \infty }{lim}{\{{(1+y)}^{-a}+\sin y\}}^{1/y}$

$=\underset{x\to \infty }{lim}\exp \left[\log {\{{(1+y)}^{-a}+\sin y\}}^{1/y}\right]$

$=\exp \left[\underset{x\to \infty }{lim}\frac{\log \{{(1+y)}^{-a}+\sin y\}}{y}\right]$

$=\exp \left[\underset{y\to 0}{lim}\frac{\frac{-a{(1+y)}^{-a-1}+\cos y}{\{{(1+y)}^{-a}+\sin y\}}}{1}\right]$ ....... [Using L'Hospital's rule]

$=e^{1-a}$
Hence, option 'B' is correct.