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

lim_{x→0} x^{2} sin(1/x) = 0.

We will now find the limit of x^{2} sin(1/x) using the Sandwich/Squeeze theorem of limits.

## Lim_{x→0} x^{2} 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 x^{2} ≥ 0, it follows that

-x^{2} ≤ x^{2} sin(1/x) ≤ x^{2}.

Now taking the limit lim_{x→0} on both sides, we obtain that

lim_{x→0} (-x^{2}) ≤ lim_{x→0} x^{2} sin(1/x) ≤ lim_{x→0} x^{2}.

⇒ 0 ≤ lim_{x→0} x^{2} sin(1/x) ≤ 0.

So by the Squeeze theorem, we get that lim_{x→0} x^{2} sin(1/x) = 0.

Thus, the limit of x^{2} sin(1/x) is equal to 0 when x tends to 0, and this is proved using the Squeeze theorem. That is,

lim_{x→0} x^{2} sin(1/x) = 0 |

More Limits:

Limit of (1+ 1/x)^{x} when x→∞

## FAQs

### Q1: What is the limit of x^{2}sin(1/x) as x approaches 0?

Answer: The limit of x^{2}sin(1/x) as x approaches 0 is equal to 1.