Yet another limit question in this forum.

[Ballox]

Homework Statement

limx->0+ [tan($$\sqrt{x}$$)-sin($$\sqrt{x}$$)] / x$$\sqrt{2x}$$

Homework Equations

tanx= sinx/cosx

The Attempt at a Solution

I set u= $$\sqrt{x}$$
Therefore as a->0, u->0

And I get the limit
=lim x->0+ (tan($$\sqrt{x}$$)-sin($$\sqrt{x}$$))/ (x*$$\sqrt{2}$$* $$\sqrt{x}$$)
= 1/$$\sqrt{2}$$lim u->0 (tan(u)-sin(u))/u³
= 1/$$\sqrt{2}$$lim u->0 (sin(u)-sin(u)cos(u))/ (cos(u)*u³)
= 1/$$\sqrt{2}$$lim u->0 sin(u)(1-cos(u))/(cos(u)*u³)

And it is here where I'm stuck. I'm not sure what to do next =\
We haven't learned L'hopital's rule yet, so if it's needed, it really shouldn't be applied.

I'm open to any suggestions and I thank you for your time,

Ballox

 

[micromass]

Do you know the limit $$\lim_{x\rightarrow 0}{\frac{sin(x)}{x}}=1$$, if yes, you can apply this two times here...

 

[Ballox]

Hmm. I do know that. I'll see what I can do.
Thanks for the hint.

 

[micromass]

Alright. Say something if it still doesn't work

 

[Ballox]

Do you know the limit $$\lim_{x\rightarrow 0}{\frac{sin(x)}{x}}=1$$, if yes, you can apply this two times here...

Alright, so I just did it again...
And in fact I was able to apply that theorem THREE times.

My final answer was 1/2$$\sqrt{2}$$

Would this answer be correct?

 

[micromass]

That looks about right

 

[Ballox]

That looks about right

Thank you kindly for your help.