Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 15

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PROBLEM:

Evaluate lim⁡x→0(sin⁡3xsin⁡x−tan⁡x) \displaystyle \lim_{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right).

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SOLUTION:

A straight substitution of x=π4\displaystyle x=\frac{\pi }{4} leads to the indeterminate form 00\displaystyle \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→0(sin⁡3xsin⁡x−tan⁡x)=lim⁡x→0(sin⁡3xsin⁡x−sin⁡xcos⁡x)=lim⁡x→0(sin⁡3xsin⁡xcos⁡x−sin⁡xcos⁡x)=lim⁡x→0(sin⁡3xcos⁡xsin⁡xcos⁡x−sin⁡x)=lim⁡x→ 0(sin⁡3xcos⁡x(sin⁡x)(cos⁡x−1))=lim⁡x→0(sin⁡2xcos⁡x(cos⁡x−1))=lim⁡x→0((1−cos⁡2x)⋅cos⁡x−(1−cos⁡x))=lim⁡x→0((1+cos⁡x)(1−cos⁡x)⋅cos⁡x−(1−cos⁡x))=lim⁡x→0((1+cos⁡x)⋅cos⁡(x)−1)=−1⋅lim⁡x→0((1+cos⁡x)⋅cos⁡x)=−1⋅(1+cos⁡0)⋅cos⁡0=−1⋅(1+1)⋅1=−2  (Answer)\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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