# Question #bc73c

Oct 17, 2016

We know that $f \left(0\right) = 1$,

so the function is continuous at $0$ if ${\lim}_{x \rightarrow 0} f \left(x\right) = 1$.

I don't have a way to evaluate this limit without l'Hopital's Rule, so I hope you have it to work with. ( See edit below. )

${\lim}_{x \rightarrow 0} \arctan x = 0$ and ${\lim}_{x \rightarrow 0} \sin x = 0$,

So the initial form of ${\lim}_{x \rightarrow 0} \arctan \frac{x}{\sin} x$ is $\frac{0}{0}$.

This is an indeterminate form to which we can apply l'Hopital's Rule.

${\lim}_{x \rightarrow 0} \arctan \frac{x}{\sin} x = {\lim}_{x \rightarrow 0} \frac{\frac{1}{1 + {x}^{2}}}{\cos} x = \frac{\frac{1}{1 + 0}}{1} = 1$.

Yes, $f$ is continuous at $0$.

Edit I do have a way to evaluate this without l"Hopital

${\lim}_{x \rightarrow 0} \arctan \frac{x}{x} = {\lim}_{h \rightarrow 0} \frac{\arctan \left(0 + h\right) - \arctan \left(0\right)}{h}$

With $f \left(t\right) = \arctan x$, this is $f ' \left(0\right)$

$\frac{d}{\mathrm{dt}} \arctan t = \frac{1}{1 + {t}^{2}}$, so $f ' \left(0\right) = 1$

Therefore ${\lim}_{x \rightarrow 0} \arctan \frac{x}{x} = 1$.

Back to the main question

${\lim}_{x \rightarrow 0} \arctan \frac{x}{\sin} x = {\lim}_{x \rightarrow 0} \left(\arctan \frac{x}{x} \cdot \frac{x}{\sin} x\right)$

$= {\lim}_{x \rightarrow 0} \left(\arctan \frac{x}{x}\right) \cdot {\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} x\right)$

$= \left(1\right) \left(1\right) = 1$