Evaluate $\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}$ without L'Hospital

Question

$$\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}$$

How can I evaluate this limit without using the L'Hospital Rule? I've expanded $\sin(4x)$ as $\sin(2x+2x)$, $\sin(3x) = \sin(2x + x)$, but none of these things worked.

Answer 1

The simplest way is to note $\sin x \simeq x$ for small $x$ and $\cos x \simeq 1-\frac{x^2}{2}$ for small $x$. Then, you can obtain $$\frac{1-\cos\sin 4x}{\sin^2\sin 3x} \simeq \frac{1-\cos 4x}{\sin^2 3x} \simeq \frac{8x^2}{9x^2} = \frac{8}{9}.$$ You need to use Taylor series to formalize this type of argument.

Answer 2

You can start by writing

$${1-\cos(\sin(4x))\over\sin^2(\sin(3x))}= {1-\cos(\sin(4x))\over\sin^2(4x)} \left({\sin(4x)\over4x}\right)^2 \left({4x\over3x}\right)^2 \left({3x\over\sin(3x)}\right)^2 \left({\sin(3x)\over\sin(\sin(3x))}\right)^2$$

Now note that

$$\lim_{x\rightarrow0}{1-\cos(\sin(4x))\over\sin^2(4x)}=\lim_{u\rightarrow0}{1-\cos u\over u^2}={1\over2}$$

$$\lim_{x\rightarrow0}{\sin(4x)\over4x}=\lim_{u\rightarrow0}{\sin u\over u}=1$$

and so forth for the others. This leads to

$$\lim_{x\rightarrow0}{1-\cos(\sin(4x))\over\sin^2(\sin(3x))}={1\over2}\left(1\right)^2\left({4\over3}\right)^2(1)^2(1)^2={8\over9}$$

At the very least you can write

$$\lim_{x\rightarrow0}{1-\cos(\sin(4x))\over\sin^2(\sin(3x))}={16\over9}\lim_{u\rightarrow0}{1-\cos u\over u^2}\lim_{u\rightarrow0}{u\over\sin u}$$

(assuming the two limits on the right hand side exist).

Answer 3

Start with a Taylor's polynomial. $cos x = 1 -x^2/2! + o(x^4)$. $sinx = x -x^3/3! +o(x^6)$. The o's just mean that the remaining terms are like $x^4$ and $x^6$. For purposes of seeing what happens at x = 0 they are irrelevant. With this we have $sin^2x = x^2 +o(x^4).$

So $(1-cosx)/sin^2(x) \approx (x^2/2)/x^2 = 1/2. \hspace{50px}$(1)

Since you were asked about sin3x and sin4x let's put those into (1).

$(1-cos(sin4x)/sin^2(sin3x) \approx [(sin^2(4x))/2]/sin^23x$.

By the computations above $sin^24x \approx (4x)^2 + o(x^4)$ and $sin^23x \approx (3x)^2 + o(x^4)$ . Putting it together we have

$[(sin^2(4x))/2]/sin^23x \approx [(4x)^2/2]/(3x)^2 =[16x^2/2]/9x^2 = 8/9$