How do you use the Squeeze Theorem to find #lim xcos(50pi/x)# as x approaches zero?

Question

Elijah Abram · Accepted Answer

See the explanation.

From trigonometry #-1 <= cos theta <=1# for all real numbers #theta#. So, #-1 <= cos ((50pi)/x) <= 1# for all #x != 0# From the right For #x > 0#, we can multiply without changing the directions of the inequalities, so we get: #-x <= cos ((50pi)/x) <= x# for #x > 0#. Observe that , #lim_(xrarr0^+) (-x) = lim_(xrarr0^+) (x) = 0#, so, #lim_(xrarr0^+)cos ((50pi)/x) = 0# From the left For #x < 0#, when we multiply we must change the directions of the inequalities, so we get: #-x >= cos ((50pi)/x) >= x# for #x < 0#. Observe that , #lim_(xrarr0^-) (-x) = lim_(xrarr0^-) (x) = 0#, so, #lim_(xrarr0^-)cos ((50pi)/x) = 0# Dual-sided Limit Because both the left and rights limits are #0#, we conclude that: #lim_(xrarr0)cos ((50pi)/x) = 0#

Elijah Abram · Answer

undefined To use the Squeeze Theorem to find lim xcos(50pi/x) as x approaches zero, we can start by noting that -1 ≤ cos(50pi/x) ≤ 1 for all x ≠ 0.

Next, we can multiply each part of the inequality by x to get -x ≤ xcos(50pi/x) ≤ x for all x ≠ 0.

Since the limit of -x as x approaches zero is 0, and the limit of x as x approaches zero is also 0, we can conclude that the limit of xcos(50pi/x) as x approaches zero is also 0.