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

hint
After multiplying by the conjugate as you done, the denominator becomes
$$1+x\sin(x)-\cos(x)=$$
$$2\sin(\frac x2)\Bigl(x\cos(\frac x2)+\sin(\frac x2)\Bigr)$$
The function can written as
$$\frac{x}{2\sin(\frac x2)}\frac{1}{\cos(\frac x2)+\frac 1x\sin(\frac x2)}$$
$$×(\sqrt{1+x\sin(x)}+\sqrt{\cos(x)})$$
the limit is then
$$1×\frac{1}{1+\frac 12}×2=\frac 43$$
