# Solving Limit Problem without L'Hospital Rule $\lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}$

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

• Use $1-\cos\theta=2\sin^2\frac\theta2$ Mar 2, 2022 at 2:09
• 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. Mar 2, 2022 at 2:12
• $\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 $\ \lim_{x \ \rightarrow \ 0} \ \frac{\sin x}{x} \ = \ 1 \ \ .$ Mar 2, 2022 at 20:09

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?

• You are missing a minus sign at the end.
– Gary
Mar 2, 2022 at 2:21
• @Gary thanks for pointing it out. I have fixed it. Mar 2, 2022 at 2:22

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