# How to use L'Hopital's rule to evaluate a limit as $x$ approaches $0$

$$\lim_{x\to 0} \ \frac{8x(\cos9x-1)}{\sin5x-5x}$$

How can do I break down the equation to find the limit?

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The title asks for implementation of l'Hopital. Did you try using l'Hopital? What happened? – Gerry Myerson Oct 22 '12 at 22:20

You do not need to use l'Hospital's rule. Recall the following: $$\cos(9x) - 1 = -2 \sin^2(9x/2)$$ $$\lim_{\theta \to 0} \dfrac{\sin(\theta)}{\theta} = 1$$ $$\lim_{\theta \to 0} \dfrac{\sin(\theta) - \theta}{\theta^3} = -\dfrac16$$

Hence, $$\dfrac{8x}{\sin(5x) - 5x} (\cos(9x)-1) = \dfrac{8x^3}{\sin(5x) - 5x} \dfrac{\cos(9x)-1}{x^2} = \dfrac{8x^3}{\sin(5x) - 5x} \dfrac{-2 \sin^2(9x/2)}{x^2}$$ $$\dfrac{8x}{\sin(5x) - 5x} (\cos(9x)-1) = - \dfrac{16}{5^3} \dfrac{(5x)^3}{\sin(5x) - 5x} \times \left(\dfrac{9}2 \right)^2 \left(\dfrac{\sin(9x/2)}{(9x/2)} \right)^2$$ Now make use of the known limits to get the answer.

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Sorta begs the question, how one knows that $-1/6$ limit; especially, how or whether OP knows it. – Gerry Myerson Oct 22 '12 at 22:31
@GerryMyerson Have added the appropriate link now. – user17762 Oct 22 '12 at 22:35

$$\lim_{x \to 0} \frac{8x \cos(9x) - 8x}{\sin(5x) - 5x} = \lim_{x \to 0} \frac{8 \cos(9x) - 72x\sin(9x) - 8}{5 \cos(5x) - 5} = \lim_{x \to 0} \frac{-72\sin(9x) - 72\sin(9x) - 72(9)x\cos(9x)}{-25 \sin(5x)} = \lim_{x \to 0} \frac{-72(9)(2)\cos(9x) - 72(9)\cos(9x) + 72(81)x\sin(9x) }{-125 \cos(5x)} = \frac{3(9)(72)}{125} =$$

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Hey, here's an idea --- why not do Don's homework for him? – Gerry Myerson Oct 22 '12 at 22:29

Using l'Hospital's rule we have $$\mathop {\lim }\limits_{x \to 0} \frac{{8x(\cos (9x) - 1)}}{{\sin (5x) - 5x}} = \mathop {\lim }\limits_{x \to 0} \frac{{8\cos (9x) - 8 - 72x\sin (9x)}}{{5\cos (5x) - 5}}$$ so ...

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Should the term after the 72 be a sin instead? – Amzoti Oct 22 '12 at 22:26
Yes. Thank you. – glebovg Oct 22 '12 at 22:32

$$f(x)=8x(\cos{9x}-1)$$ and $$g(x)=\sin{5x}-5x.$$ So : $$\lim_{x \to 0}{\frac{f(x)}{g(x)}}=\lim_{x \to 0}{\frac{f'(x)}{g'(x)}}=\frac{8(\cos{9x}-1)+72x(-\sin{9x})}{5\cos{5x}-5}=\lim_{x \to 0}{\frac{f''(x)}{g''(x)}}=\lim_{x \to 0}\frac{-72\sin{9x}-72\sin{9x}-9\cdot72x\cdot\cos{9x}}{-25\sin{5x}}=\lim_{x\to 0}\frac{-144\sin{9x}-9\cdot 72x \cdot \cos{9x}}{-25\sin{5x}}=\lim_{x \to 0}{\frac{144}{25}\cdot \frac{\sin{9x}}{\sin{5x}}}+\lim_{x \to 0}{\frac{9\cdot 72}{25}\cdot\frac{x\cdot \cos{9x}}{\sin{5x}}=\frac{144}{25}\cdot \frac{9}{5}+\lim_{x\to 0}{\frac{9 \cdot72}{25}\cdot\frac{\cos{9x}-9x\sin{9x}}{5\cos{5x}}}}=\frac{144 \cdot 9}{125}+\frac{9 \cdot 72}{125}=\frac{9 \cdot 216}{125}.$$ I hope is all right :) It's all about calculus.

The most important part is :

$$\lim_{x \to 0}{\frac{f(x)}{g(x)}}=\lim_{x \to 0}{\frac{f'(x)}{g'(x)}}$$

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I hope it's all wrong. The problem is labelled "homework". Why are you doing Don's homework for him? Why not give Don a chance to work it out on his own? Why deprive him of the joy? – Gerry Myerson Oct 22 '12 at 22:51
@GerryMyerson I guess is ok and nothing is wrong but please excuse me. I didn't see the label "homework". Next time I hope to be more careful . – Iuli Oct 22 '12 at 23:10

Let's try one line proof: we know that $\sin x=x-\frac{x^3}{6}+O(x^5)$. Then

$$\lim_{x\to 0} \ \frac{8x(\cos9x-1)}{\sin5x-5x}=\lim_{x\to 0} \left(\frac{8x}{\displaystyle\frac{125 x}{6}}\cdot \frac{1-\cos9x}{(9x)^2}\cdot81\right)=\frac{48}{125}\cdot \frac{1}{2} \cdot 81=15.552$$

Chris.

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