JEE Main 2013 (Online) 9th AprilQuestion 1
The value of $\lim_{x\to0}\frac1x\left[\tan^{-1}\left(\frac{x+1}{2x+1}\right)-\frac\pi4\right]$ is :
Limits & Continuity
Solution
Concept. At $x=0$, $\tan^{-1}(1)=\frac\pi4$, so the limit is the derivative of $F(x)=\tan^{-1}\frac{x+1}{2x+1}$ at $0$. Formulas used. (1) $\lim_{x\to0}\frac{F(x)-F(0)}{x}=F'(0)$(2) $\frac{d}{dx}\tan^{-1}u=\frac{u'}{1+u^2}$(3) quotient rule: $\big(\frac pq\big)'=\frac{p'q-pq'}{q^2}$Step 1 - $u$ and $u'$ (3). $u=\frac{x+1}{2x+1}$, $u'=\frac{(2x+1)-2(x+1)}{(2x+1)^2}=\frac{-1}{(2x+1)^2}$. At $0$: $u=1$, $u'=-1$. Step 2 - Derivative (2). $F'(0)=\frac{-1}{1+1}=-\frac12$. Step 3 - Use (1). Limit $=-\frac12$. Answer: B