# How to evaluate the limit $\lim_{x\to 0} \frac{1-\cos(4x)}{\sin^2(7x)}$

I am lost in trying to figure out how to evaluate the $$\lim_{x\to 0} \frac{1-\cos(4x)}{\sin^2(7x)}.$$

So far, I have tried the following:

Multiply the numerator and denominator by the numerator's conjugate $1+\cos(4x)$, which gives $\frac{\sin^2(4x)}{(\sin^2(7x))(1+\cos(4x))}$. However, I am not sure what to do after this step.

Could anyone please help point me in the right direction as to what I am supposed to do next?

All help is appreciated.

• Perhaps you meant to evaluate $$\lim_{x\to 0} \frac{1-\cos(4x)}{\sin^2(7x)}$$ instead? Oct 7, 2015 at 18:34
• Yes, that is what I meant to evaluate. Oct 7, 2015 at 18:34

Notice, $$\lim_{x\to 0}\frac{1-\cos(4x)}{\sin^2(7x)}$$ $$\lim_{x\to 0}\frac{1-\cos^2(4x)}{\sin^2(7x)(1+\cos (4x))}$$

$$=\lim_{x\to 0}\frac{\sin^2(4x)}{\sin^2(7x)(1+\cos (4x))}$$ $$=\lim_{x\to 0}\frac{\sin^2(4x)}{\sin^2(7x)}\cdot \lim_{x\to 0} \frac{1}{1+\cos (4x)}$$ $$=\lim_{x\to 0}\left(\frac{\sin(4x)}{\sin(7x)}\right)^2\cdot \frac{1}{1+1}$$ $$=\frac{1}{2}\lim_{x\to 0}\left(\frac{4}{7}\frac{\frac{\sin(4x)}{4x}}{\frac{\sin(7x)}{7x}}\right)^2$$ $$=\frac{1}{2}\frac{16}{49}\lim_{x\to 0}\left(\frac{\frac{\sin(4x)}{(4x)}}{\frac{\sin(7x)}{(7x)}}\right)^2$$ $$=\frac{8}{49}\left(\frac{1}{1}\right)^2$$ $$=\color{red}{\frac{8}{49}}$$

• +1. I'd like to add that intuitively, because of what we know about $\frac{\sin(x)}{x}$ and $\frac{1-\cos(x)}{x^2}$, $\sin(7x)$ is behaving like $7x$ while $1-\cos(4x)$ is behaving like $\frac{(4x)^2}{2}=8x^2$. So the ratio is behaving like $\frac{8x^2}{49x^2}$. The argument above is one way to make this precise.
– Ian
Oct 7, 2015 at 18:40

Hint:

$$\lim_{t\to 0} \frac{\sin^2(4t)}{\sin^2(7t)}=\lim_{t\to 0}\frac{\left(\frac{\sin 4t}{t}\right)^2}{\left(\frac{\sin 7t}{t}\right)^2}=\frac{4^2}{7^2}\times\frac{\left(\lim_{t\to 0}\frac{\sin 4t}{4t}\right)^2}{\left(\lim_{t\to 0}\frac{\sin 7t}{7t}\right)^2}=\frac{16}{49}\times \frac{1^2}{1^2}$$ Also \begin{align} \lim_{t\to 0}\frac{1}{1+\cos(4t)}&=\frac{1}{1+1}=\frac{1}{2} \end{align}

You could also use L'Hopital's rule,

$$\lim_{x\to 0}\frac{1-\cos(4x)}{\sin^2(7x)}$$

$$=\lim_{x\to 0}\frac{4\sin(4x)}{7\sin(14x)}$$

$$=\lim_{x\to 0}\frac{16\cos(4x)}{7\times14\cos(7x)}$$

$$=\frac{16\cos(0)}{7\times14\cos(0)} = \frac{8}{49}$$

HINT:

Using $\cos2A=1-2\sin^2A,$

$$\dfrac{1-\cos4x}{\sin^27x}=2\cdot2^2\cdot\left(\dfrac{\sin2x}{2x}\right)^2\cdot\dfrac1{7^2\cdot\left(\dfrac{\sin7x}{7x}\right)^2}$$