Evaluate a limit: lim_{x->0^+} (e^x - e^{sin(x)})/(sqrt[x] (x - tan(x)) * sin((x - 3sqrt[x])/(1 - x))

[Saul Feyn]

Hello,

I'm trying to solve this limit but I can't figure it out, could anyone be so kind to help me?


[math]\lim_{x \to 0^+} \frac{e^x-e^{\sin(x)}}{\sqrt x [x-\tan x]} \sin \frac{x-3\sqrt x}{1-x}[/math]

The correct answer should be: [math] -\frac{3}{2}[/math]

 

[Dr.Peterson]

I'm trying to solve this limit but I can't figure it out, could anyone be so kind to help me?

[math]\lim_{x \to 0^+} \frac{e^x-e^{\sin(x)}}{\sqrt x [x-\tan x]} \sin \frac{x-3\sqrt x}{1-x}[/math]
The correct answer should be: [math] -\frac{3}{2}[/math]

Please show what you've tried. Did you try making it a single fraction and applying L'Hopital's rule? (I haven't tried yet, because it will be very complicated, but it's what I'd do if I had to (though I might try doing the two factors separately, in case they have workable limits).

Where did this horror come from? Are there any other techniques you've learned, such as series?

 

[Saul Feyn]

Sorry, I opened this thread on Saturday but it needed to be verified and in the meantime I managed to solve it.

I was stuck on the [imath]\sin \frac{x-3\sqrt x}{1-x}[/imath] but I've been suggested to see it like [imath]\sin(t) \approx t, t \approx 0[/imath] and I got it. I simplified what I could and then I applied L'Hopital's rule, and it turned out to be quite simple.

It's from a Calculus 1 exam.

Thanks for your time.