Question

Question #bc73c

Answer 1

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

so the function is continuous at `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`