Question

The value of $\underset{x\to 0}{lim}[{\tan }^{-1}\left(\frac{2x+1}{4x+1}\right)-\frac{x}{4}]$ is equal to :

• A. $\frac{\pi }{4}$
• B. $-\frac{\pi }{2}$
• C. $\frac{-\pi }{4}$
• D. $-\frac{3\pi }{4}$

Answer 1

Answer: (A)

$\frac{\pi }{4}$

$\underset{x\to 0}{lim}[{\tan }^{-1}\left(\frac{2x+1}{4x+1}\right)-\frac{x}{4}]$= $\underset{x\to 0}{lim}[{\tan }^{-1}\left(\frac{2\times 0+1}{4\times 0+1}\right)-\frac{0}{4}]=ta{n}^{-1}1=\frac{\pi }{4}$