How do I find the limit as #x# approaches infinity of #xsin(1/x)#?

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Answer 1 · 2014-09-26T23:28:32

By l'Hopital's Rule,

#lim_{x to infty}x sin(1/x)=1#

Let us look at some details.

#lim_{x to infty}x sin(1/x)#

by rewriting a bit,

#=lim_{x to infty}{sin(1/x)}/{1/x}#

by l'Hopital's Rule,

#=lim_{x to infty}{cos(1/x)cdot(-1/x^2)}/{-1/x^2}#

by cancelling out #-1/x^2#,

#=lim_{x to infty}cos(1/x)=cos(0)=1#