A calculus problem by Tasneem Khaled

Let

$L=\displaystyle\lim_{x\rightarrow 1} \dfrac{\sin(x-1)}{x^3-1}$

By direct substitution we get the indeterminate form $\dfrac{0}{0}$ so we can apply L'Hôpital's Rule

$\begin{aligned} L&=\displaystyle\lim_{x\rightarrow 1}\dfrac{\dfrac{d}{dx}[\sin(x-1)]}{\dfrac{d}{dx}[x^3-1]}\\ &=\displaystyle\lim_{x\rightarrow 1}\dfrac{\cos(x-1)}{3x^2}=\boxed{\dfrac{1}{3}} \end{aligned}$

$\displaystyle \lim_{x \to 1} \frac{\sin (x-1)}{x^3-1}$

Applying L'Hôpital's Rule

$= \displaystyle \lim_{x \to 1} \dfrac{\cos(-1+x)}{3x^2}$

$= \dfrac{\cos (-1+1)}{3}$

$= \dfrac13 \approx 0.3333$