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Use $\lim_{x\to0} \frac{\sin x}{x} = 1$ to evaluate the limits:

a) $$\lim_{x\to0} \frac{x\tan^2(x)}{\cos(3x)\sin^3(2x)}$$

b) $$\lim_{x\to \frac{\pi}{2}} \frac{\tan(2x)}{x-\frac{\pi}{2}}$$

Can someone teach me how to do this please

Peter Huxford
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stacks-geek
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2 Answers2

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For part a, rewrite:

\begin{align*} \lim_{x\to0}\frac{x\tan^2(x)}{\cos(3x)\sin^3(2x)}&=\lim_{x\to0}\frac{x\sin^2(x)}{\cos(3x)\cos^2(x)\sin^3(2x)} \\ &=\lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{x^3}{\sin^3(2x)}\cdot\frac{1}{\cos(3x)\cos^2(x)} \\ &=\lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{(2x)^3}{\sin^3(2x)}\cdot\frac{1}{8\cos(3x)\cos^2(x)} \\ &=\left(\lim_{x\to0}\frac{\sin x}{x}\right)^2\cdot\left(\lim_{2x\to0}\frac{\sin(2x)}{2x}\right)^{-3}\left(\lim_{x\to0}\frac{1}{8\cos(3x)\cos^2(x)}\right) \\ &=1^2\cdot1^{-3}\cdot\frac{1}{8} \\ &=\frac{1}{8} \end{align*}

For part b: let $t=x-\frac{\pi}{2}$. Then as $x\to\frac{\pi}{2}$, $t\to0$. So the limit becomes:

$$\lim_{t\to0}\frac{\tan(2t-\pi)}{t}=\lim_{t\to0}\frac{\tan(2t)}{t}=\lim_{t\to0}\frac{\sin(2t)}{t\cos(2t)}=\left(\lim_{t\to0}\frac{\sin(2t)}{2t}\right)\left(\lim_{t\to0}\frac{2}{\cos(2t)}\right)$$

For the first factor, let $u=2t$. Then $t\to0\implies u\to0$, so this limit becomes:

$$\lim_{x\to\frac{\pi}{2}} \frac{\tan(2x)}{x-\frac{\pi}{2}}=\left(\lim_{u\to0}\frac{\sin(u)}{u}\right)\left(\lim_{t\to0}\frac{2}{\cos(2t)}\right)=1\cdot\frac{2}{1}=2$$

Peter Huxford
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Hints: Some useful identities,
$tan(x)=sin(x)/cos(x)$
$sin(2x)=2sin(x)cos(x)$
$cos(3x)=4cos^3(x)-3cos(x)$
$tan(\pi/2-x)=cot(x)=1/tan(x)$

The first formula equals to $$\frac{x}{8sin(x)}\cdot \frac{1}{cos^5(x)(4cos^3(x)-3cos(x))}$$

The second one, if you replace $y=\pi/2-x$ you get $tan(2y)/y$