Question

Let$r=\underset{x\to 0}{lim}{\left(\begin{array}{c}\frac{sinx}{x}\end{array}\right)}^{\frac{1}{x}}$and$m=\underset{x\to 0}{lim}{\left(\begin{array}{c}\frac{sinx}{x}\end{array}\right)}^{\frac{1}{x^2}}$, then -

• A. $r=0andm=e^{1/6}$
• B. $r=1andm=e^{1/6}$
• C. $r=0andm=e^{-1/6}$
• D. $r=1andm=e^{-1/6}$

Answer 1

The correct option is D $r=1andm=e^{-1/6}$

$r=\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x}}=\underset{x\to 0}{lim}exp\left(\frac{\log \left(\frac{\sin x}{x}\right)}{x}\right)$
Applying L'Hospital's rule
$r=exp\left(\underset{x\to 0}{lim}\frac{x\cos x-\sin x}{x\sin x}\right)$
Applying L'Hospital's rule
$r=exp\left(\underset{x\to 0}{lim}-\frac{\sin x}{(x\cos x+\sin x)}\right)=e^0=1$
And
$m=\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}=\underset{x\to 0}{lim}exp\left(\frac{\log \left(\frac{\sin x}{x}\right)}{x^2}\right)$
Applying L'Hospital's rule
$m=exp\left(\underset{x\to 0}{lim}\frac{x\cos x-\sin x}{2x^2\sin x}\right)$
Applying L'Hospital's rule
$m=exp\left(\underset{x\to 0}{lim}-\frac{\sin x}{2(x\cos x+2\sin x)}\right)$
Applying L'Hospital's rule
$m=exp\left(\underset{x\to 0}{lim}-\frac{\cos x}{2(3\cos x-x\sin x)}\right)=e^{-\frac{1}{6}}$