Need help solving a trig limit

**xxlvh** · November 3rd 2009, 05:23 PM

Evaluate the following limit without the use of L'Hopital's rule:
$\lim_{x\to0}\frac{sin^2x}{\sqrt{1+xsinx}-cosx}$

My first instinct was to multiply by the conjugate, leaving me with:
$\lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{1+x sinx-cos^2x}$

And from here I tried rearranging as best I could but could not end up with any form without zero in the denominator.

**alexmahone** · November 3rd 2009, 05:49 PM

Originally Posted by xxlvh

Evaluate the following limit without the use of L'Hopital's rule:
$\lim_{x\to0}\frac{sin^2x}{\sqrt{1+xsinx}-cosx}$

My first instinct was to multiply by the conjugate, leaving me with:
$\lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{1+x sinx-cos^2x}$

And from here I tried rearranging as best I could but could not end up with any form without zero in the denominator.

$\lim_{x\to0}\frac{sin^2x}{\sqrt{1+xsinx}-cosx}$

= $\lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{1+x sinx-cos^2x}$

= $\lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{sin ^2x+xsinx}$

= $\lim_{x\to0}\frac{\sqrt{1+xsinx}+cosx}{1+\frac{x}{ sinx}}$

= $\frac{\sqrt{1+0}+1}{1+1}$

= $\frac{2}{2}$

= $1$

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