How do you find the limit of #x*sin(1/x)# as x tends to positive infinity?

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Answer 1 · 2015-05-20T15:39:46

You can use L'Hopital's Rule to say the answer is 1:

#lim_{x->infty}xsin(1/x)=lim_{x->infty}\frac{sin(1/x)}{1/x}#

#lim_{x->infty}\frac{-x^{-2}cos(1/x)}{-x^{-2}}=lim_{x->infty}cos(1/x)=cos(0)=1#.

Answer 2 · 2015-05-20T16:35:14

If you prefer to find the limit without using l'Hopital, rewrite:

#lim_(xrarroo) xsin(1/x) = lim_(xrarroo)(sin(1/x)/(1/x))#

Let #theta = 1/x# and observe that as #x rarr oo#, #theta rarr0#.

So
#lim_(xrarroo) xsin(1/x) = lim_(theta rarr 0)(sin theta/theta) = 1#