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

The limit of x sin(1/x) as x approaches 0 is equal to 0. This limit is denoted by limx→0 xsin(1/x). So the formula for the limit of x sin(1/x) when x tends to zero is as follows.

limx→0 x sin(1/x) = 0.

Let us now find the limit of xsin(1/x) using the Squeeze/Sandwich theorem.

## Proof of Limx→0 xsin(1/x) =0 by Squeeze Theorem

Explanation:

Step 1:

From the Trigonometry we know that sinθ lies between -1 and 1 for any real values of θ. Thus we have that

-1 ≤ sin(1/x) ≤ 1.

Taking modulus, we get that

0 ≤ |sin(1/x)| ≤ 1.

Step 2:

Multiplying both sides by |x|, it follows that

0 ≤ |x| |sin(1/x)| ≤ |x|

⇒ 0 ≤ |xsin(1/x)| ≤ |x|.

Step 3:

Taking limx→0 on both sides, we get

limx→0 (0) ≤ limx→0 |x sin(1/x)| ≤ limx→0 |x|.

⇒ 0 ≤ limx→0 |x sin(1/x)| ≤ 0.

Thus, by the Squeeze theorem, limx→0 |x sin(1/x)| = 0.

As we know that limx→0|f(x)| = 0 ⇒ limx→0 f(x) = 0, it follows that limx→0 xsin(1/x) = 0.

Conclusion: So the limit of xsin(1/x) is equal to 0 when x tends to 0, and this is obtained by the Squeeze theorem. That is,

More Limits:

Limit of x2sin(1/x) when x→0

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 xsin(1/x) as x approaches 0?

Answer: The limit of xsin(1/x) as x approaches 0 is equal to 1, that is, limx→0 x sin(1/x) =0.