Question

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

Answer 1

See the explanation section.

Explanation:

If we have been through the proof of `lim_(thetararr0)sintheta/theta = 1`, then we have seen that for small, positive `theta`, we have:

`0 < sintheta < theta`.

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

So, for large positive `x`, we have

`0 < sin(1/x) < 1/x`.

Observe that `sqrtx> 0`, so we can multiply without reversing the inequalities.

`0 < sqrtx sin(1/x) < 1/sqrtx`

`lim_(xrarroo)0 = 0 = lim_(xrarroo)1/sqrtx`.

Therefore,

`lim_(xrarroo)sqrtxsin(1/x) = 0`.

Note
Here is a picture reminder for `0 < theta < pi/2` that `0 < sintheta < 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`.

`sintheta` 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 < sintheta < theta`