Question

For $a\in R^{+}$ such that $\underset{x\to 0}{lim}\frac{1-\cos ax}{x^2}$$=\underset{x\to \pi }{lim}\frac{\sin x}{\pi -x}$

• A. $1$
• B. $\sqrt{2}$
• C. $2$
• D. none of these

Answer 1

The correct option is B $\sqrt{2}$

$\underset{x\to 0}{lim}\frac{1-\cos ax}{x^2}=\frac{0}{0}$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore, $\underset{x\to 0}{lim}\frac{\frac{d}{dx}(1-\cos ax)}{\frac{d}{dx}\left(x^2\right)}=\underset{x\to 0}{lim}\frac{a\sin ax}{2x}=\frac{0}{0}$ which is again an indeterminate form.

Hence, L-hospital's rule is applied once again.

Therefore, $\underset{x\to 0}{lim}\frac{\frac{d}{dx}\left(a\sin ax\right)}{\frac{d}{dx}\left(2x\right)}=\underset{x\to 0}{lim}\frac{a^2\cos ax}{2}=\frac{a^2\left(1\right)}{2}=\frac{a^2}{2}$ ------(1)

$\underset{x\to \pi }{lim}\frac{\sin x}{\pi -x}=\frac{0}{0}$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule

Hence, $\underset{x\to \pi }{lim}\frac{\frac{d}{dx}\left(\sin x\right)}{\frac{d}{dx}(\pi -x)}=\underset{x\to \pi }{lim}\frac{\cos x}{-1}=\frac{-1}{-1}=1$ -------(2)

Given that, $\underset{x\to 0}{lim}\frac{1-\cos ax}{x^2}=\underset{x\to \pi }{lim}\frac{\sin x}{\pi -x}$

$⟹\frac{a^2}{2}=1$ (from (1) and (2))

$⟹a^2=2$

$⟹a=\sqrt{2}$