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.)
![enter image source here]()
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
