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Answer 1 · 2016-10-17T11:56:17

We know that $f (0) = 1$ $f(0)=1$ ,

so the function is continuous at $0$ $0$ if $lim_(xrarr0)f(x) = 1$ .

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. )

$lim_(xrarr0)arctanx=0$ and $lim_(xrarr0)sinx = 0$ ,

So the initial form of $lim_(xrarr0)arctanx/sinx$ is $0/0$ .

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

$lim_(xrarr0)arctanx/sinx = lim_(xrarr0)(1/(1+x^2))/cosx = (1/(1+0))/1 = 1$ .

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

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

$lim_(xrarr0)arctanx/x = lim_(hrarr0)(arctan(0+h)-arctan(0))/h$

With $f(t) = arctanx$ , this is $f'(0)$

$d/dt arctant = 1/(1+t^2)$ , so $f'(0)=1$

Therefore $lim_(xrarr0)arctanx/x = 1$ .

Back to the main question

$lim_(xrarr0) arctanx/sinx = lim_(xrarr0) (arctanx/x*x/sinx)$

$= lim_(xrarr0) (arctanx/x) * lim_(xrarr0)(x/sinx)$

$= (1)(1)=1$