# How to prove limit as x approaches 0 of x sin(1/x)= 0 ?

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49
Oct 11, 2015

${\lim}_{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) = 0$

#### Explanation:

Here's an answer using the squeeze or sandwich theorem.

We know that any sine must be between -1 or 1, or in mathematical terms,

$- 1 \le \sin \left(\theta\right) \le 1$

If $\theta = \frac{1}{x}$ then

$- 1 \le \sin \left(\frac{1}{x}\right) \le 1$

Multiplying both sides by $x$

$- x \le x \sin \left(\frac{1}{x}\right) \le x$ for $x > 0$
$- x \ge x \sin \left(\frac{1}{x}\right) \ge x$ for $x < 0$

And since

${\lim}_{x \rightarrow {0}^{+}} x = {\lim}_{x \rightarrow {0}^{+}} - x = 0$
${\lim}_{x \rightarrow {0}^{-}} x = {\lim}_{x \rightarrow {0}^{-}} - x = 0$

We can use the squeeze theorem to say

${\lim}_{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) = 0$

Then teach the underlying concepts
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#### Explanation

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#### Explanation:

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42
Nov 28, 2016

${\lim}_{x \to 0} x \sin \left(\frac{1}{x}\right) = 0$

#### Explanation:

Here we are, one year later with a different way to solve this in case somebody comes across this:

${\lim}_{x \to 0} x \sin \left(\frac{1}{x}\right) = {\lim}_{x \to 0} \sin \frac{\frac{1}{x}}{\frac{1}{x}}$

If we say $u = \frac{1}{x}$ then when $x \to 0$, $u \to \infty$ so we have

${\lim}_{u \to \infty} \sin \frac{u}{u}$

Remember that the sine is always a number between -1 and 1, so when $u$ grows without bound, the denominator becomes much too large and the whole fraction becomes closer and closer to zero. In other words:

${\lim}_{u \to \infty} \sin \frac{u}{u} = 0$

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