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

Question

Emma Aldridge · Accepted Answer

We need to consider right and left limits separately.

For all #x != 0#, we have #-1 <= cos(sin(1/x)) <= 1# Right Limit For #0 < x < pi/2#, we have #tanx > 0#, so we can multiply the three parts of the inequality above by #tanx# without changing the inequalities. For #0 < x < pi/2#, we get #-tanx <= tanxcos(sin(1/x)) <= tanx# Since #lim_(xrarr0^+)(-tanx )=0=lim_(xrarr0^+)(tanx )#, we have (by the right hand squeeze theorem) #lim_(xrarr0^+)tanxcos(sin(1/x)) =0# Left Limit For #-pi/2 < x < 0#, we have #tanx < 0#, so when we multiply the three parts of the inequality by #tanx# we must change the inequalities. For #-pi/2 < x < 0#, we get #-tanx >= tanxcos(sin(1/x)) >= tanx# Since #lim_(xrarr0^-)(-tanx )=0=lim_(xrarr0^-)(tanx )#, we have (by the right hand squeeze theorem) #lim_(xrarr0^-)tanxcos(sin(1/x)) =0# Unbalanced Cap Because both the right and left limits at #0# are #0#, we conclude: #lim_(xrarr0)tanxcos(sin(1/x)) =0#

Isabella Ackerman · Answer

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

First, we observe that -1 ≤ sin(1/x) ≤ 1 for all x ≠ 0. Therefore, -cos(1/x) ≤ cos(sin(1/x)) ≤ cos(1/x) for all x ≠ 0.

Next, we know that -1 ≤ cos(1/x) ≤ 1 for all x ≠ 0.

Since -1 ≤ -cos(1/x) ≤ cos(sin(1/x)) ≤ cos(1/x) ≤ 1 for all x ≠ 0, we can apply the Squeeze Theorem.

As x approaches zero, both -1 and 1 approach zero. Therefore, by the Squeeze Theorem, the limit of cos(sin(1/x)) as x approaches zero is also zero.

Finally, we can use the fact that the limit of tan(x) as x approaches zero is also zero. Therefore, the limit of tan(x)cos(sin(1/x)) as x approaches zero is zero.