# How to find $\lim_{x\to 0}\frac{\tan 3x}{\tan 5x}$?

I am asked to find the following limit:

$$\lim_{x \to 0} \frac{\tan 3x}{\tan 5x}$$

My problem is in simplifying the function. I followed two different approaches to solve the problem. But both seems incorrect.

Apprach 1) Since $\tan \theta = \frac{sin \theta}{cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we have:

$$\frac{\tan 3x}{\tan 5x} = \frac{\sin 3x \times \cos 5x}{\sin 5x \times \cos 3x}$$

This approach does not work well, because I cannot simplify more.

Approach 2) Since $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \times \tan y)}$, we have:

$$\frac{\tan 3x}{\tan 5x} = \frac{\tan 3x}{\tan 3x + \tan 2x} = \frac{\tan 3x}{\frac{\tan 2 + \tan3}{1 - \tan 2 \times \tan 3}}$$

What am I doing wrong?

• Have you got some mistakes in second approach? tan(5x)=tan(3x)+tan(2x) is not that good – k99731 Aug 22 '14 at 5:01
• @k99731 It's (hopefully) a typo, it should read $\tan (3x + 2x)$ and in the last part of that equation the denominator should be $\tan 2x$ and $\tan 3x$ not $\tan 2$ and $\tan 3$. – MCT Aug 22 '14 at 5:05
• Yah, it's a duplicate, but this post has way more answers (maybe the other question ought to be closed...). – colormegone Aug 22 '14 at 6:32

Using L'Hopital's:

$$\lim \limits_{x \to 0} \frac{\tan 3x}{\tan 5x} = \lim \limits_{x \to 0} \frac{3 \sec^2 3x}{5 \sec^2 5x} = \frac{3}{5}$$

Without L'Hopital's:

$$\lim \limits_{x \to 0} \frac{\tan 3x}{\tan 5x} = \frac{3}{5} \lim \limits_{x \to 0} \frac{\cos 5x}{\cos 3x} \cdot \frac{\sin 3x}{3x} \cdot \frac{5x}{\sin 5x} = \frac{3}{5}$$

This uses the fact that $\lim \limits_{u \to 0} \frac{\sin u}{u} = 1$. This approach derives the identities $\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac ab$ and $\lim_{x \to 0} \frac{\tan ax}{\tan bx} = \frac ab$.

Even simpler. Rewrite $$\frac{\tan 3x}{\tan 5x}= \frac{\tan 3x}{3x}\times\frac{5x}{\tan 5x}\times\frac{3}{5}$$ and remember than, for small values of $y$, $\tan(y) \simeq y$.

I am sure that you can take from here.

Note that $$\lim_{x \rightarrow 0} \frac{\sin(3x)}{\sin(5x)} = \lim_{x \rightarrow 0} \frac{3\cos(3x)}{5\cos(5x)} = \frac{3}{5}$$ by L'Hopital's rule. You can use this fact in approach 1 since $\cos(3x) \rightarrow 1$ and $\cos(5x) \rightarrow 1$.

Details:

\begin{align} \lim_{x \rightarrow 0} \frac{\sin(3x) \cos(5x)}{\sin(5x)\cos(3x)} &= \left(\lim_{x \rightarrow 0} \frac{\sin(3x)}{\sin(5x)}\right) \left(\lim_{x \rightarrow 0} \frac{\cos(5x)}{\cos(3x)}\right) \\ &= \left(\frac{3}{5}\right) \left(\frac{1}{1}\right) \\ &= \frac{3}{5} \\ \end{align}

• @Baqer: I added some more details. – Bungo Aug 22 '14 at 5:14

Approach $1)$

$$a\frac{\lim_{x\to0}\dfrac{\sin ax}{ax}}{\lim_{x\to0}\cos ax}=a\cdot\frac11$$

So, $$\lim_{x\to0}\frac{\sin ax}{\sin bx}=\frac ab\lim_{x\to0}\dfrac{\sin ax}{ax}\frac1{\lim_{x\to0}\dfrac{\sin bx}{bx}}=\frac ab\cdot\frac11$$

You can directly use L'Hopital's rule, as Bungo and Michael T have given an answer.

I just want to give an alternative answer involving simplifying $\sin(2x)$ and $\sin(3x)$.

From approach 1, we can see we only have to find $\lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)}$.

By $\sin(x+y) = \sin(x)\cos(y)+\sin(y)\cos(x)$, we can break sin term and keep cosine term to obtain the following form:

$\frac{\sin(3x)}{\sin(5x)} = \frac{2\sin(x)\cos(x)\cos(x)+\sin(x)\cos(2x)}{2\sin(x)\cos(x)\cos(x)\cos(2x)+\sin(x)\cos(2x)\cos(2x)+2\sin(x)\cos(x)\cos(3x)}$

Eliminating $\sin(x)$, and evaluate the limit at 0, we have 3/5.

If the only "trigonometric limit law" you have available is $\ \lim_{u \rightarrow 0} \ \frac{\sin \ u}{u} \ = \ 1 \$ , you can take the expression apart as

$$\lim_{x \to 0} \ \ \frac{\tan 3x}{\tan 5x} \ = \ \lim_{x \to 0} \ \ \frac{\sin 3x}{\cos 3x} \ \cdot \ \frac{\cos 5x}{\sin 5x}$$

$$= \ \lim_{x \to 0} \ \ \sin 3x \ \cdot \ \frac{3x}{3x} \ \cdot \ \frac{1}{\cos 3x} \ \cdot \ {\cos 5x} \ \cdot \ \frac{5x}{5x} \ \cdot \ \frac{1}{\sin 5x}$$

$$= \ \lim_{x \to 0} \ \frac{\sin 3x}{3x} \ \cdot \ \frac{1}{\cos 3x} \ \cdot \ \cos 5x \ \cdot \ \frac{5x}{\sin 5x} \ \cdot \ \frac{3x}{5x}$$

$$= \ \lim_{x \to 0} \ \frac{\sin 3x}{3x} \ \cdot \ \lim_{x \to 0} \ \frac{1}{\cos 3x} \ \cdot \ \lim_{x \to 0} \ \cos 5x \ \cdot \ \lim_{x \to 0} \ \frac{5x}{\sin 5x} \ \cdot \ \frac{3}{5}$$

$$= \ \lim_{3x \to 0} \ \frac{\sin 3x}{3x} \ \cdot \ \lim_{x \to 0} \ \frac{1}{\cos 3x} \ \cdot \ \lim_{x \to 0} \ \cos 5x \ \cdot \ \lim_{5x \to 0} \ \frac{1}{(\sin 5x) / 5x} \ \cdot \ \frac{3}{5}$$

$$= \ 1 \ \cdot \ 1 \ \cdot \ 1 \ \cdot \ \frac{1}{\lim_{5x \to 0} \ [ (\sin 5x) / 5x]} \ \cdot \ \frac{3}{5} \ = \ \frac{1}{1} \ \cdot \ \frac{3}{5} \ = \ \frac{3}{5} \ \ .$$

We can generalize such arguments to produce limit laws such as $\ \lim_{x \rightarrow 0} \ \frac{\sin \ ax}{\sin \ bx} \ = \ \frac{a}{b} \$ and $\ \lim_{x \rightarrow 0} \ \frac{\tan \ ax}{\tan \ bx} \ = \ \frac{a}{b} \$ .

(Reading over all the responses, this is basically the fully-detailed argument of what lab bhattacharjee shows, which is what you'd likely have to write out in an exam or homework problem early in first-semester calculus. It "builds character" and makes you grateful for L'Hopital...)