Question #5d46e

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Answer 1 · 2017-02-09T18:11:50

The limit is #0#. Use the squeeze theorem.

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

For #x > 0# and near #0#, we have #sinx > 0# so we can multiply the inequality by #sinx# to get

#-sinx <= sinx sin(1/x) <= sinx# for #0 < x <+ pi/2#

Observe that #lim_(xrarr0^+) -sinx = 0# and lim_(xrarr0^+) -sinx =0#,

therefore, #lim_(xrarr0^+) sinxsin(1/x) = 0#

For #x < 0# and near #0#, we have sinx < 0#, so when we multiply the inequality we must revese the directions:

#-sinx >= sinxsin(1/x) >= sinx#

Observe that #lim_(xrarr0^-) -sinx = 0# and lim_(xrarr0^-) -sinx =0#,

therefore, #lim_(xrarr0^-) sinxsin(1/x) = 0#.

Since both one-sided limits are #0#, the two-sided limit is also #0#.