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

Question

Ezra Adams · Accepted Answer

See the explanation section.

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#

Isaiah Allen · Answer

undefined To use the Squeeze Theorem to find the limit of sqrt(x)sin(1/x) as x approaches infinity, we need to find two functions that "squeeze" the given function and have the same limit as x approaches infinity.

First, we can observe that -1 ≤ sin(1/x) ≤ 1 for all x. Therefore, we have -sqrt(x) ≤ sqrt(x)sin(1/x) ≤ sqrt(x) for all x.

Next, we can find the limits of the two bounding functions as x approaches infinity.

The limit of -sqrt(x) as x approaches infinity is -∞.

The limit of sqrt(x) as x approaches infinity is ∞.

Since -∞ ≤ sqrt(x)sin(1/x) ≤ ∞ for all x, and the limits of the bounding functions are -∞ and ∞, respectively, we can conclude that the limit of sqrt(x)sin(1/x) as x approaches infinity is also ∞.