How to solve this limit without the L'Hospital?

How would one evaluate the following limit without using L'Hospital Rule

$$\lim_{x\to -1}\dfrac{\sin(x+1)}{x^3+1}$$

the result should be $1/3$.

Answer

Hint: Use

$$\dfrac{\sin(x+1)}{x^3+1}=\dfrac{\sin(x+1)}{(x+1)(x^2-x+1)}$$

and

$$\lim_{u\to 0} \dfrac{\sin u }{u}=1.$$