12. limits — MATHS STD 11 Science — Question

Let

$p =\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{6 / x^2}$

$\Rightarrow \quad \log p =\lim _{x \rightarrow 0} \frac{6}{x^2} \log \left(\frac{x}{\sin x}\right)$

$\Rightarrow \quad \log p =\lim _{x \rightarrow 0} \frac{6 \log \left(\frac{x}{\sin x}\right)}{x^2}$

Apply $L-$ Hospital rule

$\log p=\lim _{x \rightarrow 0}$ 

$6 \frac{\frac{\sin x}{x} \frac{(\sin x-x \cos x)}{\sin ^2 x}}{2 x}$

$\log p=\lim _{x \rightarrow 0} 6 \frac{\sin x(\sin x-x \cos x)}{x \cdot 2 x \sin ^2 x}$

$\log p=\lim _{x \rightarrow 0} 3 \frac{\sin x}{x} \times \lim _{x \rightarrow 0}$

$\frac{\sin x-x \cos x}{\frac{\sin ^2 x}{x^2}} \times x^3$

$\log p=3 \times \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^3}$

$\Rightarrow \quad \log p=3 \times \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^3}$

Again apply L.Hospital's rule

$\log p=3 \lim _{x \rightarrow 0} \frac{\cos x-\cos x+x \sin x}{3 x^2}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

$\therefore \quad p=e^1=e$