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

Question

Scarlett Allison · Accepted Answer

#lim_(xrarroo)(1/x)cosx = 0#

#-1 <= cosx <=1# for all #x#.

For # x > 0#, we have #1/x > 0#, so we can multiply the inequality by #1.x# without reversing its direction.

#-1/x <= (1/x)cosx <= 1/x# for #x > 0#.

#lim_(xrarroo)(-1/x) = 0# and #lim_(xrarroo)(1/x) = 0#.

Accordingly, at infinity, by the squeeze theorem,

#lim_(xrarroo)(1/x)cosx = 0#

William Abdalla · Answer

undefined To use the Squeeze Theorem to find the limit of (1/x)cosx 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 ≤ cosx ≤ 1 for all values of x. Therefore, we can say that -1/x ≤ (1/x)cosx ≤ 1/x.

Now, let's find the limits of the two bounding functions as x approaches infinity.

As x approaches infinity, -1/x approaches 0.

Similarly, as x approaches infinity, 1/x also approaches 0.

Therefore, by the Squeeze Theorem, since -1/x ≤ (1/x)cosx ≤ 1/x and both -1/x and 1/x approach 0 as x approaches infinity, the limit of (1/x)cosx as x approaches infinity is also 0.