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

2 Answers
Jan 4, 2018

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

Explanation:

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

Jan 5, 2018

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

Explanation:

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