# What is the limit `lim_(h->0) (sin(pi+h)-sin pi)/h`

### 2 Answers | Add Yours

The value of `lim_(h->0) (sin(pi+h) - sin pi)/h` has to be determined.

`lim_(h->0) (sin(pi+h) - sin pi)/h`

=> `lim_(h->0) (sin(pi+h) - 0)/h`

=>` lim_(h->0) (sin pi*cos h + cos pi*sin h - 0)/h`

=> `lim_(h->0) (0*cos h + (-1*sin h) - 0)/h`

=> `lim_(h->0) (-1*sin h)/h`

=>` -1*lim_(h->0) sin h/h`

=> -1

**The value of `lim_(h->0) (sin(pi+h) - sin pi)/h = -1` **

For a function f(x), the limit `lim_(h->0)(f(x+h) - f(x))/h` is the derivative f'(x) of the function.

Here, we are supposed to determine `lim_(h->0) (sin(pi+h) - sin pi)/h`

`lim_(h->0) (sin(pi+h) - sin pi)/h` is the derivative of sin x for `x = pi` . There is no need to actually calculate the value of the limit, use the derivative of the function sin x which is cos x.

This gives `lim_(h->0) (sin(pi+h) - sin pi)/h = cos pi = -1`

The required limit `lim_(h->0) (sin(pi+h) - sin pi)/h = -1`