2
$\begingroup$

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?

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ Why $x^2$ disappeared in first $lim$? $\endgroup$
    – J.Olufsen
    Nov 17, 2013 at 17:46
  • $\begingroup$ because $x^2+x=x(x+1)$ $\endgroup$
    – math110
    Nov 17, 2013 at 17:50
  • $\begingroup$ I cannot get the same result. Can You give more detailed solution? $\endgroup$
    – J.Olufsen
    Nov 17, 2013 at 18:02
  • $\begingroup$ How can You transform $(x+1)$ multiplied by sum of radicals to $2$? $\endgroup$
    – J.Olufsen
    Nov 17, 2013 at 18:09
  • 1
    $\begingroup$ $\lim_{x\to 0}(x+1)=1$ and $\lim_{x\to 0}(\sqrt{1-\sin{(5x)}}+\sqrt{1+\sin{(5x)}})=2$.... $\endgroup$
    – math110
    Nov 17, 2013 at 18:16
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .