Using L'Hopital's Rule

Use L'Hopital's Rule to evaluate the limit $\lim_{x{\rightarrow0}}\frac{x \cos(x)}{x+\arcsin(x)}$

Here arcsin: (-1,1)-->(-pi/2,pi/2) is the inverse function of the sine function.

If I let $f(x)=x\cos(x)$ and $g(x)=x+\arcsin(x)$, we obviously know that $f$ is differentiable but how do we know that $g$ is differentiable?

Since the sum of two differentiable functions is differentiable over their common domain, it's only sufficient to know arcsin(x) is differentiable.

Can you assume that? I'm sure you've covered it at A-level.

Since the sum of two differentiable functions is differentiable over their common domain, it's only sufficient to know arcsin(x) is differentiable.

Can you assume that? I'm sure you've covered it at A-level.

Inverse Function Theorem makes it differentiable in a neighbourhood of 0, anyway.

I know what the derivative of arcsinx is but our lecturer is veryyy uptight with justification so I really doubt she'd let us assume it. Is there any way to say arcsinx is differentiable?

And no... we didn't do anything to do with derivative of arcsinx at alevel.
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Have you covered the "Inverse Function Theorem", as Smaug123 suggested - basically dy/dx = 1/(dx/dy)

Can't see how you'd justify it without using that.

I did try but the inverse f thm has closed intervals and the derivative is never equal to zero.

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If you need it to be closed, you can restrict the domain over which arcsin is defined - as long as it includes 0, there's no problem.

But if it includes 0 then the derivative would itself be 0 which makes a problem doesnt it? The thm says the derivative can't equal 0

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But if it includes 0 then the derivative would itself be 0 which makes a problem doesnt it? The thm says the derivative can't equal 0

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