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!
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 $$|f'(x)|\le \frac{2}{x}+\frac{1}{x^2}\le 3,\quad x\ge 1.$$