# Find the limit of $\frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) }$ without L'Hospital technique

I would like to find the limit as $x$ goes to zero for the following function, but without L'Hospital technique.

$$f(x) = \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)}$$

This limit will go to zero (I had tried using calculator manually).

I have tried to open the trigonometric identities and the fact that $\lim\limits_{x \rightarrow 0}\frac{\sin (ax)}{x} = a$ gives : $$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\cos (3x)( \cos (3x) - 1)}{6 x^{2}}$$ $$\cos (3x) = \cos (2x) \cos(x) - \sin (2x) \sin(x) = \cos^{3}(x) - 3\sin^{2}(x) \cos(x)$$ $$\lim_{x \rightarrow 0 } \frac{ \cos (3x) }{x^{2}} = \lim_{x \rightarrow 0} \frac{\cos^{3}(x)}{x^{2}} - 3$$

Any view on this and more efficient way to solve this? Thanks before.

• You have serious issue in your last equation. It is much easier to see that factor $\cos 3x$ tends to $1$ and hence the desired limit is equal to limit of $(\cos 3x - 1)/6x^{2}$. Now multiply and divide by $1+\cos 3x$ and you should get the answer easily. – Paramanand Singh Sep 19 '17 at 19:42

$$\lim_{x\to0}\dfrac{\cos3x-1}{\sin2x\tan3x}=-\lim_{x\to0}\dfrac{\sin^23x\cos3x}{\sin2x\sin3x(1+\cos3x)}$$

$$=-\dfrac32\cdot\lim_{x\to0}\dfrac{\sin3x}{3x}\cdot\dfrac1{\lim_{x\to0}\dfrac{\sin2x}{2x}}\cdot\lim_{x\to0}\dfrac{\cos3x}{1+\cos3x}=?$$

You have to recall three basic limits: \begin{align} &\lim_{t\to0}\frac{\sin t}{t}=1 \\[6px] &\lim_{t\to0}\frac{\tan t}{t}=1 \\[6px] &\lim_{t\to0}\frac{1-\cos t}{t^2}=\frac{1}{2} \end{align} In your problem, the coefficients of $x$ can be dealt with easily: $$\frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)} = \frac{ \cos (3x) - 1 }{(3x)^2}\frac{2x}{\sin(2x)}\frac{3x}{\tan(3x)} \frac{3^2}{2\cdot 3}$$ where the last (numeric) fraction has been inserted to keep the expression the same.

Now compute the limit of each fraction and you're done.

\begin{align} \lim_{x\to0} \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)} &= \lim_{x\to0} \frac{ -2\sin^2 (\dfrac{3x}{2}) \cos (3x) }{ \sin (2x) \sin (3x)} \\ &= -2\lim_{x\to0} \dfrac{\sin^2 (\dfrac{3x}{2}) }{(\dfrac{3x}{2})^2} \dfrac{(2x)}{\sin (2x)} \frac{ (3x)}{ ( \sin (3x)}.\dfrac{(\dfrac{3x}{2})^2}{(2x)(3x)}\cos (3x) \\ &=\color{blue}{-\dfrac{3}{4}} \end{align}

$$\lim_{x\rightarrow0}\frac{\cos 3x - 1}{ \sin 2x \tan 3x }=-\lim_{x\rightarrow0}\frac{2\sin^2\frac{3x}{2}}{\sin2x\sin3x}=$$ $$=-\frac{3}{4}\lim_{x\rightarrow0}\frac{\frac{\sin^2\frac{3x}{2}}{\frac{9x^2}{4}}}{\frac{\sin2x}{2x}\cdot\frac{\sin3x}{3x}}=-\frac{3}{4}$$

Just another way using the standard Taylor expansions.

$$\cos(3x)=1-\frac{9 x^2}{2}+\frac{27 x^4}{8}+O\left(x^6\right)$$ $$\sin(2x)=2 x-\frac{4 x^3}{3}+O\left(x^5\right)$$ $$\tan(3x)=3 x+9 x^3+O\left(x^5\right)$$ makes $$\cos(3x)-1=-\frac{9 x^2}{2}+\frac{27 x^4}{8}+O\left(x^6\right)$$ $$\sin(2x)\tan(3x)=6 x^2+14 x^4+O\left(x^6\right)$$ Now, long division leads to $$\frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) }=-\frac{3}{4}+\frac{37 x^2}{16}+O\left(x^3\right)$$ showing the limit and how it is appraoched.

Just for illustration, use $x=\frac \pi {12}$ (quite far away from $0$); the exact value of the expression is $\sqrt{2}-2\approx -0.585786$ while the above approximation would give $\frac{37 \pi ^2}{2304}-\frac{3}{4}\approx -0.591504$.

The thinking/attempt in the question post is quite complicated. Simple idea related to @ParamanandSingh 's comment. Multiply the function by $\frac{\cos(3x)+1}{\cos(3x)+1}$, we get $$f(x) = \frac{\cos^{2}(3x) - 1}{\sin(2x) \tan(3x)(\cos(3x)+1)} =\frac{- \sin(3x) \cos(3x)}{\sin(2x) (\cos(3x)+1)} = \frac{- (2x) \sin(3x) \cos(3x)}{(3x) \sin(2x) (\cos(3x)+1)} \left( \frac{3}{2} \right)$$ As $x \rightarrow 0$, the limit will go to $$\frac{-1}{2} \left(\frac{3}{2} \right) = -\frac{3}{4}$$ Thanks for the inputs on this question.