Find the limit as x approaches infinity of #xsin(1/x)#?

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Wataru Share
Mar 1, 2018

By l'Hopital's Rule, we can find
#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 little bit,
#=lim_{x to infty}{sin(1/x)}/{1/x}#

by l'Ho[ital'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#

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Jim H Share
Apr 8, 2015

Instead of l'Hopital's Rule, one can use the fundamental trigonometric limit: #lim_(hrarr0)sin h/h=1#.

The limit you are interested in can be written: #lim_(xrarroo)sin (1/x)/(1/x)#.

Now, as #xrarroo#, we know that #1/xrarr0# and we can think of the limit as

#lim_(1/xrarr0)sin (1/x)/(1/x)#.

With #h=1/x#, this becomes#lim_(hrarr0)sin h/h# which is #1#.

Although it is NOT needed, here's the graph of the function:

graph{y = x sin(1/x) [-5.55, 5.55, -2.775, 2.774]}

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Jun 6, 2015

When you substitute in infinity, #oo#, you end up with the indeterminate form of #oo*0#.

#lim_(x->oo) xsin(1/x)=oo*sin(1/oo)=oo*sin(0)=oo*0#

We still have options though. We now can fall back on L'Hopital's Rule which basically says to take the derivative of the numerator and denominator independently. Do not use the quotient rule.

We need to rewrite this function so that is produces an indeterminate in the form #oo/oo# or #0/0#.

#lim_(x->oo) (sin(1/x))/x^(-1)=sin(1/x)/(1/x)=sin(1/oo)/(1/oo)=sin(0)/(0)=0/0#

Applying L'Hopital

#lim_(x->oo)((sin(1/x))')/((x^(-1))')#

#=lim_(x->oo)((-1)*x^(-2)*cos(1/x))/((-1)*x^(-2))#

Simplify the previous step

#=lim_(x->oo)cos(1/x)=cos(1/oo)=cos(0)=1#

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