<s>[INST] Define continuity for $f(x)=\arctan(2x^3)/x^2$ at $x=0$. $$f(x)=\dfrac{\arctan(2x^3)}{x^2}.$$


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*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 :)
 [/INST] 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 $(1)$.
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