# Find $\lim_{x\to0}\frac{\sin5x}{\sin4x}$ using $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$.

I am trying to find $$\lim_{x\to0}\frac{\sin5x}{\sin4x}$$

My approach is to break up the numerator into $4x+x$. So,

$$\begin{equation*} \lim_{x\to0}\frac{\sin(4x+x)}{\sin4x}=\lim_{x\to0}\frac{\sin4x\cos x+\cos4x\sin x}{\sin4x}\\ =\lim_{x\to0}(\cos x +\cos4x\cdot\frac{\sin x}{\sin4x})\end{equation*}$$

Now the problem is with $\frac{\sin x}{\sin4x}$. If I use the double angle formula twice, it is going to complicate the problem.

The hint says that you can use $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$.

I have little clue how can I make use of the hint.

Any helps are greatly appreciated. Thanks!

• Your approach will not lead you to use the given limit because whatever substitution you make, you will remain stuck with fractions involving trigonometric functions only (and not $x$ alone). – Yves Daoust Jul 30 '14 at 9:17
• – Martin Sleziak Sep 24 '16 at 7:59

Hint: Try to insert some useful parts in the given fraction properly. I mean $$\frac{5x}{4x}$$ Note that we can do this cause while $x\to 0$ then $x\neq0$.

Hint:

$$\lim_{x\to 0} \frac{\sin 5x}{\sin 4x} = \frac{5}{4}\lim_{x\to 0} \frac{\sin 5x}{5x}·\frac{4x}{\sin 4x}$$

Actually it won't complicate the problem. You can get the same result by using formula

$$\sin(2x)= 2\sin(x)\cos(x)$$

i.e. in the denominator of second term expand $\sin(4x)$ as $$2\sin(2x)\cos(2x) = 4\sin(x)\cos(x)\cos(2x).$$ By cancelling sin(x) in numerator and denominator you get a whole function in cosine terms. Just substitute zero and you get the limit.

• I must say this is a very smart solution and it shows that the limit of $(\sin 4x)/(\sin 5x)$ can be calculated without using the result that $(\sin x)/x \to 1$ as $x \to 0$. +1 – Paramanand Singh Jul 30 '14 at 10:54

$$\lim_{x\to0}\frac{\sin5x}{\sin4x}=\lim_{x\to0}\frac{5x\frac{\sin5x}{5x}}{4x\frac{\sin4x}{4x}}=\frac{5}{4}\frac{\lim_{x\to0}\frac{\sin5x}{5x}}{\lim_{x\to0}\frac{\sin4x}{4x}}=\frac{5}{4}$$

Applying the hint will need some trigonometric manipulations. Instead, the shortest possible way to find is to apply L'Hospital's Rule. So by that you have $\lim_{x\to 0} \frac{5 \cos 5x}{4 \cos 4x}$, in which case the limit $\frac{\cos 5x}{\cos 4x}$ goes to $1$ and are you are left with $\frac{5}{4}$.

$$\lim_{x\to0}{\sin5x\over\sin4x}=\lim_{x\to0}{\sin5x\over5x}{4x\over\sin4x}\frac54.$$ $$\lim_{x\to0}{\sin5x\over5x}=\lim_{x\to0}{4x\over\sin4x}=1,$$ thus we are done.

we have to find $\lim_{x\to0}\frac{\sin5x}{\sin4x}$. hence$$\lim_{x\to0}\frac{\sin5x}{\sin4x} = \frac{5}{4}\lim_{x\to0}\frac{\sin 5x}{5x}\cdot\frac{4x}{\sin 4x} =\frac{5}{4}\cdot\lim_{x\to0}\frac{\sin 5x}{5x}\cdot\lim_{x\to0}\frac{4x}{\sin 4x}$$ hope you got the idea.....

Firstly, there is a limit law that allows you to do "substitution" of limits:

Suppose you know that $$\lim_{x\to a} f(x) = L$$ Also suppose that $g$ is some function such that $g(x) \to a$ when $x \to b$.

Then: $$\lim_{x \to b} f(g(x)) = L$$

You can think of this as performing the substitution $t = g(x)$: $$t = g(x) \quad \text{so } x \to b \implies t \to a\\ \lim_{x \to b} f(g(x)) = \lim_{t \to a}f(t) = L$$

The upshot of this is that all of these limits are equal to 1: $$\lim_{x\to 0} \frac{\sin{6x}}{6x} = \lim_{x \to 0} \frac{\sin(3x^2)}{3x^2} = \lim_{x \to 1} \frac{\sin(x-1)}{x-1} = \lim_{x\to 0} \frac{\sin(\sin(x))}{\sin(x)} =\lim_{x\to 0} \frac{\sin(x)}{x} = 1$$ And, of course, innumerably many like them.

Knowing that all of these versions of the fundamental trig limit give the same result means that if we can arrange for our limit to include any of them, then that part of the limit will be done for us. This advice should help for other limits you might see in the future.

So we can do the following: \begin{align} \lim_{x\to 0} \frac{\sin(5x)}{\sin(4x)} &= \lim_{x\to 0} \frac{\sin(5x)}{1}\frac{1}{\sin(4x)}\\ &= \lim_{x\to 0} \frac{5\sin(5x)}{5x} \frac{4x}{4\sin(4x)}\\ &= \lim_{x\to 0} \frac{5}{4}\frac{\sin(5x)}{5x}\frac{4x}{\sin(4x)}\\ &= \frac{5}{4} \times 1 \times 1\\ &=\frac54 \end{align}

What I wanted to point out is that if $x\to0$, then $4x$ and $5x\to0$ as well.

Other than that earth shattering revelation, the rest of my answer is pretty much the same as most of the other posts.

$$\begin{array}{lll} \lim_{x \to 0}\frac{\sin5x}{\sin4x}&=&\lim_{x \to 0}\frac{\frac{\sin5x}{1}}{\frac{\sin4x}{1}}\\ &=&\lim_{x \to 0}\frac{\frac{\sin5x}{1}}{\frac{\sin4x}{1}}\cdot\frac{\frac{5x}{5x}}{\frac{4x}{4x}}\\ &=&\lim_{x \to 0}\frac{\frac{\sin5x}{5x}}{\frac{\sin4x}{4x}}\cdot\frac{\frac{5x}{1}}{\frac{4x}{1}}\\ &=&\lim_{x \to 0}\frac{\frac{\sin5x}{5x}}{\frac{\sin4x}{4x}}\cdot\frac{5x}{4x}\\ &=&\frac{\displaystyle\lim_{x \to 0}\frac{\sin5x}{5x}}{\displaystyle\lim_{x \to 0}\frac{\sin4x}{4x}}\cdot\lim_{x \to 0}\frac{5x}{4x}\\ &=&\frac{1}{1}\cdot\frac{5}{4}\\ &=&\frac{5}{4}\\ \end{array}$$ Notice how on the second line I essentially multiplied by 1 (e.g. $1=5x/5x=4x/4x$).