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$$f(x)=\dfrac{\arctan(2x^3)}{x^2}.$$

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

  2. 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 :)

Hakim
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mathgrad93
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  • Note that the whole expression is absolute value. – mathgrad93 Oct 19 '14 at 14:13
  • Hint: What can you say about $$\lim_{x\to0^+}\dfrac{\arctan(2x^3)}{x^2}\quad\color{grey}{\text{and}}\quad{}{}{}{}{}{}{}{}{}\lim\limits_{x\to0^-}\dfrac{\arctan(2x^3)}{x^2}.$$ – Hakim Oct 19 '14 at 14:14

3 Answers3

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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 $(1)$.

Michael Hardy
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Eclipse Sun
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  • This is the worst MathJax coding I've ever seen, and I've seen some atrocious stuff. You've got \underset{x\to 0}{\mathop{\lim }}, where you could just write \lim_{x\to 0} and lots of similarly bad things. For example, \left({{x}^{3}}\right) where you could have (x^3). ${}\qquad{}$ – Michael Hardy Oct 19 '14 at 14:44
  • @MichaelHardy I used Mathtype to help me produce a formula. Yes, it is often awkward. – Eclipse Sun Oct 19 '14 at 14:49
  • Thanks! However, isn't f(0) a 0/0 case? Or can we say it is defined due to the two sided limit? – mathgrad93 Oct 19 '14 at 14:58
  • @mathgrad93 The given expression is not defined at $x = 0$, and we are to determine a value at $x = 0$, so that the function is continuous on the whole real line. That is, just compute the (two sided) limit. – Eclipse Sun Oct 19 '14 at 15:06
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Take $$\lim_{x\to 0} \frac{\arctan (2x^3)}{x^2}$$

to find what happens when $x$ aproaches zero both sides.

$\hskip1.5in$enter image description here

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 $x \to 0$?

$\hskip1.5in$enter image description here

Michael Hardy
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Aaron Maroja
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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.

Michael Hardy
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