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

1 Answer
Mar 31, 2018

#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#