Continuity and derivability of $(\int_{x}^{1}\frac{dt}{1-\cos(t^8)})^{-1}$ Let $f(x):[0,+\infty[\to\mathbb{R}$ be the following function. Determine if $f(x)$ is continuous and derivable in $x=0$
$$f(x)= \left\{ \begin{array}{l}
\dfrac{1}{\displaystyle\int_{x}^{1}\dfrac{dt}{1-\cos(t^8)}} & \text{if }\, x>0\\[2ex] 
0 &\text{if }\,x=0 \end{array} \right. 
$$
What we have to do is $\lim_{x \to 0^+}f(x)$ and $\underset{x\to 0 }{\mathop{\lim }} \frac{f(x)-f(0)}{x-0}=\underset{x\to 0 }{\mathop{\lim }} f(x)$.
Could Taylor work for this problem or are there other solutions?

By Lagrange remainder formula,
$$1-\cos t=\frac{t^2}2+\frac{\cos k}{4!}t^4$$ for some $k\in(0,t)$.
So, $$1-\cos t^8=\frac{t^{16}}2+ \frac{\cos k}{4!}t^{32}$$
Effortlessly, $$\frac{t^{16}}2\le1-\cos t^8\le t^{16}$$ for $|t|\le 1$.
Thus, for $1\ge x>0$, 
$$\frac{15}2\frac{x^{15}}{1-x^{15}}\le f(x)\le 15 \frac{x^{15}}{1-x^{15}} $$
As a result, by squeeze theorem, you can easily show that
$$\lim_{x\to 0^+}f(x)=0=f(0)$$
$$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{f(x)}{x}=0$$, proving continuity and derivability (I usually call it differentiability).
