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PROBLEM:
Evaluate \displaystyle \lim_{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right).
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SOLUTION:
A straight substitution of \displaystyle x=\frac{\pi }{4} leads to the indeterminate form \displaystyle \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows
\begin{align*} \displaystyle \lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right) & =\lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\frac{\sin x}{\cos x}}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\sin^3 x}{\frac{\sin x \cos x - \sin x}{\cos x}}\right) \\ \\ & = \lim _{x\to 0}\left(\frac{\sin^3 x \cos x }{\sin x \cos x - \sin x}\right) \\ \\ &=\lim _{x\to \:0}\left(\frac{\sin^3 x \cos x}{\left(\sin x \right)\left(\cos x-1\right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\sin^2 x \cos x }{\left(\cos x -1\right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\left(1-\cos^2 x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\left(1+\cos x \right) \left(1-\cos x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\left(1+\cos x \right) \cdot \cos\left(x\right)}{-1}\right) \\ \\ & =-1\cdot \lim _{x\to 0}\left(\left(1+\cos x\right)\cdot\cos x \right) \\ \\ & =-1\cdot \left(1+\cos 0 \right)\cdot \cos 0 \\ \\ & =-1\cdot \left(1+1\right)\cdot 1 \\ \\ & =-2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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