Question

How do you find the Limit of `(x(cosx-1))/(sinx-x)` as `x` approaches `0`?

Answer 1

`lim_(x->0)(x(cosx-1))/(sinx-x)=3`

Explanation:

The problem is to evaluate

`lim_(x->0)(x(cosx-1))/(sinx-x)=lim_(x->0)(xcosx-x)/(sinx-x)`

Currently, the limit takes the form of `0/0`

Therefore, we can apply L'Hôpital's Rule.

Taking the derivative of both the numerator and denominator:

`rArrlim_(x->0)(cosx-xsinx-1)/(cosx-1)`

... which is still `0/0`

Let's try L'Hôpital again:

`rArrlim_(x->0)(-sinx-(sinx+xcosx))/(-sinx)`

`rArrlim_(x->0)(-2sinx-xcosx)/(-sinx)`

... which is still `0/0`

Using L'Hôpital one more time:

`rArrlim_(x->0)(-2cosx-(cosx-xsinx))/(-cosx)`

`rArrlim_(x->0)(-3cosx+xsinx)/(-cosx)`

Now, evaluating this limit gives

`(-3+0)/-1=3`