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How can we find the limit $\lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}$

I keep getting DNE as my answer. The answer is supposed to be 0.

My Work:

$\lim_{x\to 0} \frac{\cos(x)-1}{1}*\frac{1}{\sin(2x)}$

$\lim_{x\to 0} \frac{\cos(x)-1}{1}*\frac{1}{\sin(2x)}*\frac{2x}{2x}$

$\lim_{x\to 0} \frac{\cos(x)-1}{1}*\frac{1}{2x}$

$\lim_{x\to 0} \frac{\cos(x)-1}{2x}$

$\lim_{x\to 0} \frac{\cos(0)-1}{2(0)}$

$\frac{0}{0}$

I am unsure what I am doing wrong. This is a problem in my HW. When I solve using L'Hospital Rule I get 0 and the answer key is in agreement with it. However we are explicitly told to solve without use of L'Hospital Rule.

Thank you

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    $\begingroup$ Use $1-\cos\theta=2\sin^2\frac\theta2$ $\endgroup$
    – Andrei
    Mar 2, 2022 at 2:09
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    $\begingroup$ I have to say, nearly all the standard questions of this sort become trivial if/when we write/allow Taylor series expansions of the functions involved. The abstract, and-indeed-correct, version of L'Hopital does apply in much greater generality, but really none of the usual questions depends on that. Even then, Taylor series are legit asymptotic expansions even for non-analytic functions, etc. $\endgroup$
    – paul garrett
    Mar 2, 2022 at 2:12
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    $\begingroup$ $ \lim_{x \ \rightarrow \ 0} \ \ \frac{\cos(x)-1}{x} \ = \ 0 \ \ $ is a useful "limit law" to know: it is used in finding the derivative rules for $ \ \sin x \ $ and $ \ \cos x \ \ . $ It is also proven without much difficulty from $ \ \lim_{x \ \rightarrow \ 0} \ \frac{\sin x}{x} \ = \ 1 \ \ . $ $\endgroup$
    – boojum
    Mar 2, 2022 at 20:09

2 Answers 2

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HINT

Multiply and divide by $\cos(x) + 1$:

\begin{align*} \lim_{x\to 0}\frac{\cos(x) - 1}{\sin(2x)} & = \lim_{x\to 0}\frac{\cos^{2}(x) - 1}{2\sin(x)\cos(x)}\times\frac{1}{\cos(x) + 1}\\\\ & = \lim_{x\to 0}-\frac{\sin^{2}(x)}{2\sin(x)\cos(x)}\times\frac{1}{\cos(x) + 1}\\\\ & = \lim_{x\to 0}-\frac{\sin(x)}{2\cos(x)}\times\frac{1}{\cos(x) + 1} \end{align*}

Can you take it from here?

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    $\begingroup$ You are missing a minus sign at the end. $\endgroup$
    – Gary
    Mar 2, 2022 at 2:21
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    $\begingroup$ @Gary thanks for pointing it out. I have fixed it. $\endgroup$
    – Átila Correia
    Mar 2, 2022 at 2:22
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$$ \lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)} $$ Also you can use the Taylor series: $$ \cos(x)=\left(1-\frac{x^2}{2}+\frac{x^4}{24}-....\right),\tag{1} $$ $$ \sin(2x)=\left(2x-\frac{(2x)^3}{3!}+\frac{(2x)^5}{5!}+\frac{(2x)^7}{7!}+....\right),\tag{2} $$ Therefore, $$ \lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}=\lim_{x\to{0}}\frac{\left(1-\frac{x^2}{2}+\frac{x^4}{24}-....\right)-1}{\left(2x-\frac{(2x)^3}{3!}+\frac{(2x)^5}{5!}+\frac{(2x)^7}{7!}+....\right)}= $$ $$ =\lim_{x\to{0}}\frac{x\left(-\frac{x}{2}+\frac{x^3}{24}-....\right)}{x\left(2-\frac{(2x)^2}{3!}+\frac{(2x)^4}{5!}+\frac{(2x)^6}{7!}+....\right)}=\lim_{x\to{0}}\frac{0}{2}=0. $$ Therefore, $$ \lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}=0. $$

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