<s>[INST] Evaluate the limit $\lim_{x\rightarrow 0} \frac{\sqrt{1-\sin(5x)}-\sqrt{1+\sin(5x)}}{x^2+x}$ Trying to find 

$$\lim_{x\rightarrow 0} \dfrac{\sqrt{1-\sin(5x)}-\sqrt{1+\sin(5x)}}{x^2+x}=\lim_{x\rightarrow 0} \dfrac{(1-\sin(5x))-(1+\sin(5x))}{(x^2+x)(\sqrt{1-\sin(5x)}+\sqrt{1+\sin(5x)})}=\lim_{x\rightarrow 0} \dfrac{-2\sin(5x)}{(x^2+x)(\sqrt{1-\sin(5x)}+\sqrt{1+\sin(5x)})}$$

How to solve it?
 [/INST] Recall
$$\sin^2(a) + \cos^2(a) = 1$$
Hence,
$$1 \pm \sin(2a) = ( \cos(a) \pm \sin(a))^2$$
For small $x$, we have
$$\sqrt{1 \pm \sin(5x)} = ( \cos(5x/2) \pm \sin(5x/2))$$
Use this along with the fact that $\lim_{x \to 0} \dfrac{\sin(\alpha x)}{x} = \alpha$ to conclude the limit.
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