Calculating limit of function

Without L'Hôpital

The expansions $\sin(x) = x - \frac{x^3}{6} + O(x^5)$ and $\cos(u) = 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6)$ combine joyfully to give $$ \cos(\sin x) - \cos(x) = \left(1 - \frac{x^2}{2} + \frac{5x^4}{24}\right) - \left(1-\frac{x^2}{2} + \frac{x^4}{24}\right) + O(x^5) $$ so finally, $$ \lim_{x\to 0} \frac{\cos(\sin x) - \cos(x)}{x^4} = \frac{5}{24}-\frac{1}{24} = \frac{1}{6}. $$

Using Prosthaphaeresis Formulas,

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

So, $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin\frac{x-\sin x}2}{\frac{x-\sin x}2}\frac{\sin\frac{x+\sin x}2}{\frac{x+\sin x}2}\cdot\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}x\cdot\frac14$$

We know, $\lim_{h\to0}\frac{\sin h}h=1$

$$\text{Apply L'Hospital's Rule on }\lim_{x\to0}\frac{x-\sin x}{x^3}$$

$$\text{and we get }\lim_{x\to0}\frac{x+\sin x}x=1+\lim_{x\to0}\frac{\sin x}x$$

One approach to problems like this is to replace the numerator and denominator by functions that have the same value, derivative, second derivative, etc. (up to some point) as the originals at the limit point. For instance, to handle $\sin(x)/x$, you replace $\sin(x)$ with $f(x) = x - x^3/3!$, which agrees with $\sin$ to 3rd order at $x = 0$, i.e., $\sin(0) = f(0)$; $\sin'(0) = f'(0)$; $\sin''(x) = f''(0)$; and $sin^{(3)}(0) = f^{(3)}(0)$. So then you look at $\lim \frac{x - x^3/6}{x} = \lim 1 - x^2/6 = 1$, and you're done.

In the case of this problem, you can replace $\sin(x)$ with $x - x^3/6$ and $\cos(x)$ with $1 - x^2/2$, so that

$\cos(\sin(x)) - \cos(x))$

becomes

$1 - \frac{(x - x^3/6)^2}{2} - (1 - x^2/2)$

When you simplify that a bit, you'll get to the answer.