Limit Calculation with L'Hospital's Rule: arctanx/arcsinx rigorously at x=0

[lep11]

Homework Statement

Calculate $\lim_{x \rightarrow 0} \frac{arctanx}{arcsinx}$ 'rigoriously'.

The Attempt at a Solution

What's the best approach? L'Hospitals rule?

$\lim_{x \rightarrow 0} \frac{arctanx}{arcsinx}=\lim_{x \rightarrow 0} \frac{\sqrt{1-x^2}}{x^2+1} =1$

 

[member 587159]

Yes, since you solved it instantly with it.

 

[lep11]

What if we 'don't know' the derivatives of arcsin and arctan?

 

[member 587159]

What if we 'don't know' the derivatives of arcsin and arctan?

Then you proof what these derivatives are equal too. It can't be solved in an easier way.

 

[lep11]

Yes, since you solved it instantly with it.

I should check the conditions for l' Hospitals rule first.

Then you proof what these derivatives are equal too. It can't be solved in an easier way.

I am thinking whether I can assume we know the derivatives or begin with calculating the derivatives first?
It's kinda re-inventing the wheel though?

How about applying taylor series of arcsin and arctan?

 

[Ray Vickson]

I should check the conditions for l' Hospitals rule first.

I am thinking whether I can assume we know the derivatives or begin with calculating the derivatives first?
It's kinda re-inventing the wheel though?

How about applying taylor series of arcsin and arctan?

How would you find the Taylor series without knowing the derivatives?

Nobody is re-inventing the wheel here. If you know the derivatives (or can find them easily) then l'Hospital's rule is useful; otherwise, it does you no good. In your case you know the derivatives, so l'Hospital works like a charm.