<s>[INST] If $f(x)=\sqrt{1-\sqrt{1-x^2}}$, then prove that $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,0) \cup (0,1)$. 
If $f(x)=\sqrt{1-\sqrt{1-x^2}}$, then prove that $f(x)$ is continuous on $[-1,1]$ and differentiable on $(-1,0) \cup (0,1)$.

Please prove using $$\lim_{x\to c^+}f(x)=\lim_{x\to c^-}f(x)=f(c)$$
and
$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}$$
Thankyou
 [/INST] Using $\lim_{x\to c}f(x)=f(c)$ to check continuity, in this case, is just the same as checking that you have a composition of continuous functions, which is continuous wherever it is defined.
Since $1-x^2\ge0$ means $|x|<1$ and implies $\sqrt{1-x^2}\le1$, your function is defined (and continuous) in $[-1,1]$.
The derivative can be computed with the chain rule, keeping in mind that the derivative of $x\mapsto\sqrt{x}$ exists only for $x>0$:
$$
f'(x)=\frac{1}{2\sqrt{1-\sqrt{1-x^2}}}\frac{x}{\sqrt{1-x^2}}
$$
You have to be careful where $1-\sqrt{1-x^2}=0$ and $\sqrt{1-x^2}=0$, that is at $x=0$, $x=1$ and $x=-1$, where the formula doesn't apply.
However, since the function is continuous, you can compute the limit of the derivative; it's easy to show that
\begin{align}
\lim_{x\to-1^+}f'(x)&=-\infty\\
\lim_{x\to 1^-}f'(x)&=\infty
\end{align}
so your function is not differentiable in $-1$ and $1$. For the limit at $0$ you can write the derivative as
\begin{align}
f'(x)&=
\frac{1}{2\sqrt{1-x^2}}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}\\
&=
\frac{1}{2\sqrt{1-x^2}}\frac{x\sqrt{1+\sqrt{1-x^2}}}{\sqrt{1-(1-x^2)}}\\
&=
\frac{\sqrt{1+\sqrt{1-x^2}}}{2\sqrt{1-x^2}}\frac{x}{\sqrt{x^2}}\\
&=
\frac{\sqrt{1+\sqrt{1-x^2}}}{2\sqrt{1-x^2}}\frac{x}{\lvert x\rvert}\\
\end{align}
so that
$$
\lim_{x\to0^-}f'(x)=-\frac{\sqrt{2}}{2},\qquad
\lim_{x\to0^+}f'(x)=\frac{\sqrt{2}}{2}
$$
which proves that $f$ is not differentiable at $0$.
Here's the graph

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