JEE PYQ: Limits Question 31
Question 31 - 2019 (11 Jan Shift 1)
Let $[x]$ denote the greatest integer less than or equal to $x$. Then: $\lim_{x \to 0} \frac{\tan(\pi\sin^2 x) + (|x| - \sin(x[x]))^2}{x^2}$:
(1) does not exist
(2) equals $\pi$
(3) equals $\pi + 1$
(4) equals $0$
Show Answer
Answer: (1)
Solution
RHL: $\lim_{x\to 0^+}\frac{\tan(\pi\sin^2 x) + (x - 0)^2}{x^2} = 1 + \pi$. LHL: $\lim_{x\to 0^-}\frac{\tan(\pi\sin^2 x) + x^2 + \sin^2 x - 2x\sin x}{x^2} = \pi + 1 + 1 - 2 = \pi$. Since LHL $\neq$ RHL, limit does not exist.
BETA
