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

Question

4677 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-15T21:51:08

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#

Two-sided Limit

Because both the left and rights limits are #0#, we conclude that:

#lim_(xrarr0)cos ((50pi)/x) = 0#