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