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

lim_{x→0} x sin(1/x) = 0.

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

## Proof of Lim_{x→0} xsin(1/x) =0 by Squeeze Theorem

Answer: The limit lim_{x→0} xsin(1/x) is equal to 0. |

**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 lim_{x→0} on both sides, we get

lim_{x→0} (0) ≤ lim_{x→0} |x sin(1/x)| ≤ lim_{x→0} |x|.

⇒ 0 ≤ lim_{x→0} |x sin(1/x)| ≤ 0.

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

As we know that lim_{x→0}|f(x)| = 0 ⇒ lim_{x→0} f(x) = 0, it follows that lim_{x→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,

lim_{x→0} xsin(1/x) = 0 |

More Limits:

Limit of x^{2}sin(1/x) when x→0

Limit of (1+ 1/x)^{x} when x→∞

## 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, lim_{x→0} x sin(1/x) =0.