How do you determine the limit of #x * sin(1/x) # as n approaches #oo#?

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Answer 1 · 2016-06-05T22:16:38

Changing variable and using L'Hôpital.

In your expression there is no #n#, so I assume that you mean

#lim_(x->infty) xsin(1/x)#.

We can do the substitution #t=1/x# and write

#lim_(x->infty) xsin(1/x)=lim_(t->0) sin(t)/t#

this limit can be solved applying the rule of L'Hôpital

#lim_(t->0) sin(t)/t=lim_(t->0) (d/dtsin(t))/(dt/dt)#

#=lim_(t->0) cos(t)=1#.