# Continuity at $x=0$ of this function

Not a hard exercise:$$f(x)=\frac{1}{x^3}\cdot \int_{-x}^x \sin(4t^2) \, \text{d}t \quad \text{where} \space x\ne 0\:$$ $$f(x)=5\:;\:x=0\:$$

Checking it's continuity at $x=0$ by using L'Hospital's rule and Newton-Leibniz axiom:

\begin{align} \lim _{x\to 0}\frac{1}{x^3}\cdot \int_{-x}^x \sin(4t^2) \, \text{d}t & =\lim_{x\to 0}\frac{2\sin(4x^2)}{3x^2} \\[10pt] & =\lim_{x\to 0}\frac{2\sin(4x^2)}{8x^2}\frac{8x^2}{3x^2} \\[10pt] & =\lim_{x\to 0}\frac{\sin(4x^2)}{4x^2}\frac{8x^2}{3x^2} \\[10pt] & =\frac{8}{3}\end{align}

• What is your question? – Rocket Man Jul 15 '15 at 15:23
• Is it correct ? – Ilya.K. Jul 15 '15 at 15:32
• @graydad : Why do you consider it important to write f\left(x\right) instead of f(x)? I think the former may mislead newbies into thinking those are necessary even in contexts where the default delimiter size is appropriate. But you went to some trouble to change multiple instances of things like f(x) to f\left(x\right). ${}\qquad{}$ – Michael Hardy Jul 15 '15 at 15:36
• I don't see an error. apart from a typo in line 2... – user251257 Jul 15 '15 at 15:41
• @MichaelHardy Ah, that was unintentional. I saw the question had been edited while I was editing (what you changed). So I copied my current edit (which had the original f\left(x\right) syntax from OP leftover) and pasted that when I made my edit. My apologies. – graydad Jul 15 '15 at 16:33

## 1 Answer

In the way it is written here, you have made a minor mistake. I recommend you to pass from the first line to the third line directly and omit the x^2s.

• I dont see any mistake, can you write it ? – Ilya.K. Jul 15 '15 at 18:22
• It' s fixed now. – Deniz Sargun Jul 15 '15 at 19:55