Question

Evaluate the limit? `lim_(x rarr 0) x^2(cos(1/x)-1)`

Answer 1

`lim_(x rarr 0) x^2(cos(1/x)-1) = 0`

Explanation:

We have:

`L = lim_(x rarr 0) x^2(cos(1/x)-1)`

Let us perform a substitution:

Let `u=1/x => x =1/u`

Then as `x rarr 0 => u rarr oo`

Substituting into the limit, we get:

`L = lim_(u rarr oo) (1/u)^2(cos(u)-1)`
`\ \ = lim_(u rarr oo) (cosu-1)/u^2`
`\ \ = lim_(u rarr oo) (cosu)/u^2-1/u^2`
`\ \ = lim_(u rarr oo) (cosu)/u^2- lim_(u rarr oo)1/u^2`

Consider the second limit:

`lim_(u rarr oo)1/u^2`

This is a trivial limit as `u^2` increases without bound, thus `1/u^2 rarr 0` as `u rarr oo`, thus:

`lim_(u rarr oo)1/u^2 = 0`

Consider now, the first limit:

`lim_(u rarr oo) (cosu)/u^2`

Now, we have `-1 le cosu le 1 => -1/u^2 le cosu /u^2 le 1/u^2`, thus:

`lim_(u rarr oo)(-1/u^2) le lim_(u rarr oo)(cosu /u^2) le lim_(u rarr oo)(1/u^2)`

And using the above result we just established that:

`0 le lim_(u rarr oo)(cosu /u^2) le 0`

So then by the sandwich, or squeeze, theorem we have:

`lim_(u rarr oo)(cosu /u^2) = 0`

Thus:

`L = 0`

Answer 2

I would use the squeeze theorem.

Explanation:

`-1<=cos(1/x)<=1` `" "` for all `x != 0`

So , subtracting `1` from each part,

`-2 <= cos(1/x)-1 <= 0` `" "` for all `x != 0`

Note that `x^2 >0` `" "` for all `x != 0`, so we can multiply through without changing the inequalities.

`-2x^2 <= x^2(cos(1/x)-1 ) <= 0` `" "` for all `x != 0`

Since `lim_(xrarr0)(-2x^2) = 0 = lim_(xrarr0)0`

the squeeze theorem tells us that

`Lim_(xrarr0)x^2(cos(1/x)-1 ) = 0`