0
$\begingroup$

Find $a$ if $f(x)$ is continuous at $x=0$. Also, analyze the uniform continuity of the following function

$$f(x) = \begin{cases} \frac{\sin(x^2)}{x} & x \in ( 0 , +\infty)\\ a & x =0; \end{cases}$$

I've said that for $ f(x) $ to be continuous in $ 0 $ . The limit when $x$ approaches $0$ from right has to be equal to $a$.

$$ \lim_{x\to 0^+}\frac{\sin(x^2)}{x} =\lim_{x\to 0^+}\frac{\sin(x^2)}{x^2} \cdot x = 1 \cdot 0 = 0 $$

So $a = 0$.

To study the uniform continuity I have said that $ [0,\infty] = [0 ,1] \cup(1,\infty)\\$.

So, $[0,1]$ is a compact set. As such, any continuous function is uniform continuous. But how do I show that it is continuous in $ (1,\infty) $ ? I've verified if the function is a Lipschitz function but to no avail ?

$$f'(x) = \frac{2x\cdot \cos(x^2) - \sin (x^2)}{x^2}$$ Can I show that this function is bounded in $(1,\infty)$ ?

Any other solution is welcomed!

  • 0
    Hint. cos(x) and sin(x) are always bounded by 1 – 2017-02-03
  • 0
    Doesn't change the answers below, but if I did not miscalculate then $f'(x) = \frac{2x^2\cdot \cos(x^2) - \sin (x^2)}{x^2}$ (the factor before $\cos(x^2)$ contains $x^2$)? – 2017-02-03

1 Answers 1

1

Hint:

Show if $f:[a,\infty)$ is continuous, and $\lim_{x\to \infty}f(x)$ exists, then $f$ is uniformly continuous on $[a,\infty)$.

On the other hand, to use your approach, we observe that $|\sin(x)|\le 1,|\cos(x)|\le 1$, hence we have $$|f'(x)|\le \frac{2}{x}+\frac{1}{x^2}\le 3,\quad x\ge 1.$$

  • 0
    Hmm the second proof was so obvious yet i was unable to see it.... Thanks a lot!! – 2017-02-03
  • 0
    About the first proof. I have a theorem that states that for any continuous function in the interval $[a;b)$ if the limit towards $b$ exists it is uniform continuous. Can this be expaned to $ b = \infty$ ? If so, do you have any useful link related to the proof of the said theorem. Would be a great help! – 2017-02-03
  • 0
    @Eduard6421 I don't have a proof of that. This is a standard question for the final exam of our elementary analysis course. The idea of the proof is that: given an $\epsilon>0$, use the definition of limit to find a large enough $M>0$, such that the difference of $f$ on $[M,\infty)$ is controlled. Then consider the two intervals $[a,M+1]$ and $[M,\infty)$, select $\delta=\min(\delta_0,\frac{1}{2})$ and prove the difference of $f$ can be controlled by $\epsilon$. – 2017-02-03