JEE Main 2019 (Online) 12th April Evening ShiftQuestion 1
$\lim_{x\to0}\frac{x+2\sin x}{\sqrt{x^{2}+2\sin x+1}-\sqrt{\sin^{2}x-x+1}}$ is
Limits & Continuity
Solution
Concept. Multiply by the conjugate; then divide top and bottom by $x$. Formulas used. (1) $\sqrt A-\sqrt B=\frac{A-B}{\sqrt A+\sqrt B}$(2) $\lim_{x\to0}\frac{\sin x}{x}=1$Step 1 - Rationalise (1). $A-B=x^2+2\sin x+1-\sin^2x+x-1=x^2+2\sin x+x-\sin^2x$. Step 2 - Rewrite. $$L=\lim\frac{(x+2\sin x)\big(\sqrt A+\sqrt B\big)}{x^2+x+2\sin x-\sin^2x}$$ Step 3 - Divide top and bottom by $x$ (2). Top: $\big(1+\frac{2\sin x}{x}\big)(\sqrt A+\sqrt B)\to3\cdot(1+1)=6$. Bottom: $x+1+\frac{2\sin x}{x}-\sin x\cdot\frac{\sin x}{x}\to0+1+2-0=3$. Step 4 - Limit. $\frac63=2$. Answer: C