What is the limit of #x^2sin(1/x)# as x approaches infinity?

Question

Isabella Ackerman · Accepted Answer

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

Given: #lim_(x->oo)x^2sin(1/x)#

If we use direct substitution we'd get the indeterminate form #oo*0#

#lim_(x->oo)x^2sin(1/x)=oo^2*sin(1/oo)=oo^2*sin(0)=oo*0#

We can use L'Hospital only if we have the indeterminate form #0/0# or #oo/oo#. We can however rewrite the limit to accommodate for that.

If we rewrite the limit as...

#lim_(x->oo)sin(1/x)/(1/x^2)#

And use direct substitution we'd get the indeterminate form of #0/0#

Now we apply L'Hospital's Rule

#lim_(x->oo)sin(1/x)/(1/x^2)=lim_(x->oo)(-cos(1/x)/x^2)/(-2/x^3)#

Simplifying:

#=lim_(x->oo)(xcos(1/x))/2#

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

Now using direct substitution we'll find the limit to be...

#=1/2*oo*cos(1/oo)=1/2*oo*cos(0)=1/2*oo*1=oo#

Ezekiel Alexander · Answer

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

Use:

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

Letting #t = 1/x# we have:

#lim_(x->oo) x^2 sin(1/x) = lim_(t->0^+) 1/t^2 sin(t)#

#color(white)(lim_(x->oo) x^2 sin(1/x)) = lim_(t->0^+) 1/t^2 sin(t)#

#color(white)(lim_(x->oo) x^2 sin(1/x)) = (lim_(t->0^+) 1/t) * (lim_(t->0^+) sin(t)/t)#

#color(white)(lim_(x->oo) x^2 sin(1/x)) = lim_(t->0^+) 1/t * 1#

#color(white)(lim_(x->oo) x^2 sin(1/x)) = oo#

Adam Adkins · Answer

undefined The limit of x^2sin(1/x) as x approaches infinity is 0.