Evaluate $\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}$ without L'Hospital $$\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}$$
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.
 A: 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.
A: 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).
A: 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
$[(sin^2(4x))/2]/sin^23x \approx [(4x)^2/2]/(3x)^2 =[16x^2/2]/9x^2 = 8/9$  
A: Approach essentially similar to @BarryCipra's:
$\sin^2\alpha=\frac{1-\cos(2\alpha)}{2}$
$1^{\text{st}}$ approach:
$$2\lim_{x\to 0}\frac{\sin^2(2\sin(4x))}{\sin^2(\sin(3x))}$$
I'll continue with $2^{\text{nd}}$ approach:
$$2\cdot\frac{1-\cos(\sin(4x))}{1-\cos(2\sin(3x))}$$
$$2\lim_{x\to 0}\frac{1-\cos(\sin(4x))}{1-\cos(2\sin(3x))}=2\lim_{x\to 0}\frac{\frac{1-\cos(\sin(4x))}{\sin^2(4x)}\cdot\sin^2(4x)}{\frac{1-\cos(2\sin(3x))}{4\sin^2(3x)}\cdot4\sin^2(3x)}$$$$=2\lim_{x\to 0}\frac{\sin^2(4x)}{4\sin^2(3x)}=\frac{1}{2}\lim_{x\to 0}\frac{\frac{1-\cos(8x)}{2}}{\frac{1-\cos(6x)}{2}}=\frac{1}{2}\lim_{x\to 0}\frac{\frac{1-\cos(8x)}{64x^2}\cdot64x^2}{\frac{1-\cos(6x)}{36x^2}\cdot36x^2}=\frac{1}{2}\lim_{x\to 0}\frac{16x^2}{9x^2}=\frac{1}{2}\cdot\frac{16}{9}=\frac{8}{9}$$
A: Multiply and divide by conjugate:
$$\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}\cdot \frac{1+\cos (\sin (4x))}{1+\cos (\sin (4x))}=\frac12\lim_{x \to 0} \frac{\sin^2(\sin(4x))}{\sin^2(\sin(3x))}=\\
\frac12\lim_{x \to 0} \frac{\sin^2(4x)}{\sin^2(3x)}=\frac12\lim_{x \to 0} \frac{(4x)^2}{(3x)^2}=\frac89.$$
Note: It was used $\sin x\sim x$ for $x\to 0$.
