CUET PG 2025 — Mathematics PYQ

Continuity ki Janch ($x=0$ par)

1. Function ki value:

   $$f(0)=0$$

2. Limit ki value:

   $$\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)$$

   Kyunki $\sin \left(\frac{1}{x}\right)$ hamesha $[-1,1]$ ke beech rehta hai:

   $$\underset{x\to 0}{\lim }x\cdot (\text{oscillating value})=0\times (\text{bounded value})=0$$

3. Result:

   $$\underset{x\to 0}{\lim }f(x)=f(0)$$

Iska matlab hai ki function $x=0$ par continuous hai. Aur $x\ne 0$ ke liye ye pehle se hi continuous hai.

Answer:

$f(x)$ is continuous for all $x\in R$.