Define continuity for $f(x)=\arctan(2x^3)/x^2$ at $x=0$. $$f(x)=\dfrac{\arctan(2x^3)}{x^2}.$$


*

*How are we allowed to define $f(x)$ at $x=0$ for it to be continuous there? 

*Find the derivative for all $x$ real numbers.
I can't see this work out since $x=0$ is not defined in the denominator.
Thanks beforehand if anyone can explain :)
 A: Compute the limit $$\lim_{x\to0} f(x)=\lim_{x\to0}\frac{\arctan ( 2x^3 )}{x^2}=\lim_{x\to0}\frac{2x^3}{x^2}=0.$$ Hence we define $f(0)=0$, and $f(x)$ is continuous.
Now compute the derivative. For $x\ne 0$, we have \begin{align}
  & f'(x)=\frac{{( \arctan ( 2x^3 ))^{\prime }}\cdot x^2-\arctan( 2x^3)\cdot{ (x^2)^{\prime }}}{(x^2)^2}=\frac{\frac{6x^4}{1+4x^6}-2x\arctan(2x^3)}{x^4} \\ 
 & =\frac{6}{1+4x^2}-\frac{2\arctan ( 2x^3 )}{x^3}.  \tag1
\end{align}
And $$f'(0)= \lim_{x\to0} \frac{\frac{\arctan( 2x^3)}{x^2}-0}{x-0} =\lim_{x\to0} \frac{\arctan ( 2x^3t)}{x^3} =\lim_{x\to0} \frac{2x^3}{x^3}=2.$$
Or we can simply let $x$ tends to zero in $(1)$.
A: Take $$\lim_{x\to 0} \frac{\arctan (2x^3)}{x^2}$$
to find what happens when $x$ aproaches zero both sides. 
$\hskip1.5in$
As for the derivarive 
$$\begin{align}f(x) &= x^{-2}\arctan(2x^3) \Rightarrow \\f'(x) &= -2x^{-3}\arctan(2x^3) + x^{-2}\frac{6x^2}{1+4x^6} \\&= \frac{6}{1+4x^6}-\frac{2\ \arctan(2x^3)}{x^3} \end{align}$$
What happens when $x \to 0$? 
$\hskip1.5in$
A: The function initially defined is in fact undefined when $x=0$ because the numerator and denominator are both $0$.  The question is how to define it at $0$ so as to make it continuous at $0$.  That means that what is asked for is a function of the form
$$
g(x) = \begin{cases} f(x) & \text{if }x\ne0,\\ c & \text{if }x=0, \end{cases}
$$
and such that $g$ is continuous at $0$.  That means the number $c$ must be $\lim\limits_{x\to0}f(x)$.
A limit $\lim\limits_{x\to\bullet}\dfrac{{}\  \bullet\  {}}\bullet$ where both the numerator and the denominator approach $0$ can be $0$ or any other number or $+\infty$ or $-\infty$, depending on which functions are in the numerator and the denominator.
