Question

Question #19cb9

Answer 1

The answer is `-sin(2)approx-0.035`.

Explanation:

One way to evaluate this limit is to use L'Hosptial's Rule.

You can use L'Hospital's Rule whenever a limit is equal to `0/0` or `oo/oo`. The rule states that `lim_(x->a)f(x)/g(x)=lim_(x->a)(f'(x))/(g'(x))`. Take note that this is NOT the quotient rule. Simply take the derivative of the numerator and the derivative of the denominator separately.

Applying the rule for this problem gives:

`lim_(x->1)-sin(1+x)/1`

`d/dx(cos(1+x))=-sin(1+x)`

`d/dx(cos(2))=0` because `cos(2)` is a constant.

`d/dx(x-1)=1`

Simplifying the limit gives:

`lim_(x->1) -sin(1+x)=-sin(2)`

`approx-0.035`