Question

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

Answer 1

`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`

Answer 2

`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`