How do you use the Squeeze Theorem to show that #sqrt (x) * e^(sin(pi/x))=0# as x approaches zero?

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Answer 1 · 2015-10-11T17:47:56

#-1 <= sin(pi/x) <= 1# for all #x != 0#.

So, #e^-1 <= e^sin(pi/x) <= e^1# for all #x != 0#.
(#e^u# is an increasing function)

For #x > 0#, #sqrtx# is defined and positive, so

#sqrtx/e <= sqrtxe^sin(pi/x) <= e sqrtx#

#lim_(xrarr0^+)sqrtx/e =0# #" "# and #" "# #lim_(xrarr0^+)esqrtx =0#.

Therefore, by the Squeeze Theorem,

#lim_(xrarr0^+)sqrtxe^sin(pi/x) =0#