The function $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=\lim _{n \rightarrow \infty} \frac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}}$ is continuous for all x in

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Continuity and Differentiability

The function

f : R \to R

defined by

f (x) = lim_{n \to \infty} \frac{c o s ( 2 π x ) - x ^{2 n} s i n ( x - 1 )}{1 + x ^{2 n + 1} - x ^{2 n}}

is
continuous for all x in

Answer: (B)

Evaluate the limit by considering cases based on

∣ x ∣

For $∣ x ∣ < 1$ : $x^{2 n} \to 0$ , so $f (x) = cos (2 π x)$
For $∣ x ∣ > 1$ : $x^{2 n}$ dominates, giving $f (x) = \frac{- s i n ( x - 1 )}{x - 1}$
For $x = 1$ : Direct evaluation gives $f (1) = 1$
For $x = - 1$ : The limit gives $f (- 1) = - (sin 2 + 1)$

Checking continuity:

At $x = - 1$ : $lim_{x \to - 1^{+}} f (x) = cos (- 2 π) = 1 \neq = f (- 1)$ . Discontinuous.
At $x = 1$ : $lim_{x \to 1^{-}} f (x) = cos (2 π) = 1 = f (1)$ , but $lim_{x \to 1^{+}} f (x) = 0 \neq = 1$ . Discontinuous.

Therefore,

f

is continuous for all

x \in R - {- 1, 1}

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option