# Limit of x^2 sin(1/x) as x approaches 0

The limit of x^2 sin(1/x) as x approaches 0 is equal to 0, and it is denoted by limx→0 x2 sin(1/x) = 0. So the limit formula of x2 sin(1/x) when x tends to zero is given by

limx→0 x2 sin(1/x) = 0.

We will now find the limit of x2 sin(1/x) using the Sandwich/Squeeze theorem of limits.

## Limx→0 x2 sin(1/x) using Squeeze Theorem

We know that sinθ lies between -1 and 1 for any real values of θ. That is,

-1 ≤ sinθ ≤ 1.

So we obtain that

-1 ≤ sin(1/x) ≤ 1.

As x2 ≥ 0, it follows that

-x2 ≤ x2 sin(1/x) ≤ x2.

Now taking the limit limx→0 on both sides, we obtain that

limx→0 (-x2) ≤ limx→0 x2 sin(1/x) ≤ limx→0 x2.

⇒ 0 ≤ limx→0 x2 sin(1/x) ≤ 0.

So by the Squeeze theorem, we get that limx→0 x2 sin(1/x) = 0.

Thus, the limit of x2 sin(1/x) is equal to 0 when x tends to 0, and this is proved using the Squeeze theorem. That is,

More Limits:

limx→0 tanx/x

Limit of (1+ 1/x)x when x→∞

Limit of cos(1/x) when x→∞

Limit of cos(1/x) when x→0

## FAQs

### Q1: What is the limit of x2sin(1/x) as x approaches 0?

Answer: The limit of x2sin(1/x) as x approaches 0 is equal to 1.