How can you find the limit of this: $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$? $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$
I would start with expanding it with $* \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}} \;$ but I don't know how to progress from there.
I can't use L'Hospital's rule. I also have 3 other exercises that are like this but if I can see one solved, I think I will be able to do the other ones as well.
Thanks

I'm going to write it out.
$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$
$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}} * \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}$
$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt{1+x\sin(x)} \; + \; x^2*\sqrt{\cos(x)}}{1+x\sin(x) \; - \; \cos(x)}$
$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt{1+x\sin(x)} \; + \; x^2*\sqrt{\cos(x)}}{x^2*\frac{\sin(x)}{x} \; x^2*\frac{1- \cos(x)}{x^2}}$
$\lim\limits_{x \to \; 0} \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\frac{\sin(x)}{x} \; +\frac{1- \cos(x)}{x^2}}$
$\frac{\sqrt{1+0}+1}{1+\frac{1}{2}} = \frac{2}{\frac{3}{2}} = \frac{4}{3}$
