How can you find the limit of this: $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$?

Question

$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$

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

Answer 1

Hint:

Multiplying numerator and denominator with $\sqrt{1+x\sin x}+\sqrt{\cos x}$, you obtain $$\frac{x^2}{1+x\sin x-\cos x}\cdot\bigl(\sqrt{1+x\sin x}+\sqrt{\cos x}\mkern1.5mu\bigr). $$ You can rewrite the fraction as $$\frac 1{\cfrac{1-\cos x}{x^2}+\cfrac{\sin x}x},$$ in which each term in the denominator corresponds to a standard limit.

Answer 2

As $x\to0$, famously $\frac{\sin x}{x}\to1$ and $\frac{1-\cos x}{x^2}=\frac12$, so $\frac{x\sin x+1-\cos x}{x^2}=\frac32$ and$$\frac{x^2}{\sqrt{1+x\sin x}-\sqrt{\cos x}}=\frac{x^2}{1+x\sin x-\cos x}(\sqrt{1+x\sin x}+\sqrt{\cos x})=\frac23\cdot2=\frac43.$$

Answer 3

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 $$1×\frac{1}{1+\frac 12}×2=\frac 43$$

Answer 4

I'm going to write it out.

$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$

$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}} * \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}$

$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt{1+x\sin(x)} \; + \; x^2*\sqrt{\cos(x)}}{1+x\sin(x) \; - \; \cos(x)}$

$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt{1+x\sin(x)} \; + \; x^2*\sqrt{\cos(x)}}{x^2*\frac{\sin(x)}{x} \; x^2*\frac{1- \cos(x)}{x^2}}$

$\lim\limits_{x \to \; 0} \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\frac{\sin(x)}{x} \; +\frac{1- \cos(x)}{x^2}}$

$\frac{\sqrt{1+0}+1}{1+\frac{1}{2}} = \frac{2}{\frac{3}{2}} = \frac{4}{3}$