Question

Evaluate $\underset{x\to 0}{lim}{\left[\frac{x-1+\cos x}{x}\right]}^{\frac{1}{x}}$

• A. $e^{-1/2}$
• B. $e^{1/2}$
• C. $-e^{-1/2}/2$
• D. $e^{1/2}/2$

Answer 1

The correct option is A $e^{-1/2}$

$\underset{x\to 0}{lim}{\left[\frac{x-1+\cos x}{x}\right]}^{\frac{1}{x}}$, [$1^{\infty }$ form ]
$=\underset{x\to 0}{lim}{[1+\frac{-1+\cos x}{x}]}^{\frac{x}{-1+\cos x}.\frac{1-\cos x}{x^2}.(-1)}$
$=e^{\underset{x\to 0}{lim}\frac{2{\sin }^{2}x/2}{x^2}.(-1)}=e^{\underset{x\to 0}{lim}\frac{{\sin }^{2}x/2}{(x/2{)}^2}.(-1/2)}=e^{-1/2}$