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 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 limx→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,

limx→0 xsin(1/x) = 0

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

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