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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 $1/3$.

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    $\begingroup$ What have you attempted so far? $\endgroup$ – Mohammad Zuhair Khan May 1 at 17:01
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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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