Question

If $I_1=\underset{x\to 0}{lim}({\tan }^{-1}\pi x-{\tan }^{-1}x)\cos x$ and $I_2=\underset{x\to 0}{lim}({\tan }^{-1}\pi x-{\tan }^{-1}x)\sin x$, then $(I_1,I_2)$ is

• A. $(0,0)$
• B. $(0,1)$
• C. $(1,0)$
• D. none of these

Answer 1

Answer: (A)

$(0,0)$

$I_1={lim}_{x\to 0}({\tan }^{-1}\pi x-{\tan }^{-1}x)\cos x$

$={lim}_{x\to 0}{\tan }^{-1}\pi x\cos x-{lim}_{x\to 0}{\tan }^{-1}x\cos x$

$={lim}_{x\to 0}\frac{{\tan }^{-1}\pi x}{\pi x}\times \pi x\times \cos x-{lim}_{x\to 0}\frac{{\tan }^{-1}x}{x}\times x\times \cos x$

$=\pi {lim}_{x\to 0}\frac{{\tan }^{-1}\pi x}{\pi x}\times x\times \cos x-{lim}_{x\to 0}\frac{{\tan }^{-1}x}{x}\times x\times \cos x$

$=\pi \times 1\times 0-1\times 0=0-0=0$

$\therefore I_1=0$

$I_2={lim}_{x\to 0}({\tan }^{-1}\pi x-{\tan }^{-1}x)\sin x$

$={lim}_{x\to 0}{\tan }^{-1}\pi x\sin x-{lim}_{x\to 0}{\tan }^{-1}x\sin x$

$={lim}_{x\to 0}\frac{{\tan }^{-1}\pi x}{\pi x}\times \pi x\times \frac{\sin x}{x}\times x-{lim}_{x\to 0}\frac{{\tan }^{-1}x}{x}\times x\times \frac{\sin x}{x}\times x$

$=\pi {lim}_{x\to 0}\frac{{\tan }^{-1}\pi x}{\pi x}\times x\times \frac{\sin x}{x}\times x-{lim}_{x\to 0}\frac{{\tan }^{-1}x}{x}\times x\times \frac{\sin x}{x}\times x$

$=\pi {lim}_{x\to 0}\frac{\tan \pi x}{x}\times \frac{\sin x}{x}{x}^2-{lim}_{x\to 0}\frac{{\tan }^{-1}x}{x}\times \frac{\sin x}{x}\times x^2$

$=\pi \times 1\times 1\times 0-1\times 1\times 0$

$\therefore I_2=0-0=0$

Hence $(I_1,I_2)=(0,0)$