# 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?

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

• Why $x^2$ disappeared in first $lim$? Commented Nov 17, 2013 at 17:46
• because $x^2+x=x(x+1)$ Commented Nov 17, 2013 at 17:50
• I cannot get the same result. Can You give more detailed solution? Commented Nov 17, 2013 at 18:02
• How can You transform $(x+1)$ multiplied by sum of radicals to $2$? Commented Nov 17, 2013 at 18:09
• $\lim_{x\to 0}(x+1)=1$ and $\lim_{x\to 0}(\sqrt{1-\sin{(5x)}}+\sqrt{1+\sin{(5x)}})=2$.... Commented Nov 17, 2013 at 18:16

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.