# How do you use the Squeeze Theorem to find lim sqrt(x)sin(1/x)  as x approaches infinity?

Oct 8, 2015

See the explanation section.

#### Explanation:

If we have been through the proof of ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$, then we have seen that for small, positive $\theta$, we have:

$0 < \sin \theta < \theta$.

(Or perhaps it was pointed out in our study of trigonometry. See note, below.)

So, for large positive $x$, we have

$0 < \sin \left(\frac{1}{x}\right) < \frac{1}{x}$.

Observe that $\sqrt{x} > 0$, so we can multiply without reversing the inequalities.

$0 < \sqrt{x} \sin \left(\frac{1}{x}\right) < \frac{1}{\sqrt{x}}$

${\lim}_{x \rightarrow \infty} 0 = 0 = {\lim}_{x \rightarrow \infty} \frac{1}{\sqrt{x}}$.

Therefore,

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

Note
Here is a picture reminder for $0 < \theta < \frac{\pi}{2}$ that $0 < \sin \theta < \theta$. (This is not a rigorous proof, simply a reminder.) For the angle measuring $\theta$ radians, note the point on the unit circle associated with $\theta$.

$\sin \theta$ is the vertical distance from the point to the $x$-axis (red line segent),
while $\theta$ is the length of the arc from the point to the $x$ axis (blue arc).

Because the perpendicular distance is the shortest, we believe that, for small positive $\theta$,

$0 < \sin \theta < \theta$