1
$\begingroup$

Find the limit value

enter image description here

Here's what I did (Above)

I think I can rationalize the numerator to solve it, but I'm having trouble rationalizing numerator, when I'm usually rationalizing the denominator.

How do I rationalize the numerator? (If I'm on the right track for solution)

$\endgroup$
2
$\begingroup$

$$\begin{align}\lim_{x\to0}\frac{\sqrt{x+2}-\sqrt2}{x}&= \lim_{x\to0}\frac{\sqrt{x+2}-\sqrt2}{x}\cdot\frac{\sqrt{x+2}+\sqrt2}{\sqrt{x+2}+\sqrt2}\\ &=\lim_{x\to0}\frac{x+2-2}{x(\sqrt{x+2}+\sqrt2)}\\ &=\lim_{x\to0}\frac{x}{x(\sqrt{x+2}+\sqrt2)}\\ &=\lim_{x\to0}\frac{1}{\sqrt{x+2}+\sqrt2}\\ &=\frac{1}{2\sqrt2}\\ \end{align}$$

$\endgroup$
0
$\begingroup$

$$\lim_{x\to0}\frac{\sqrt{x+2}-\sqrt2}{x} = \lim_{x\to0}\frac{\sqrt{2+x}-\sqrt2}{x-0} = (\sqrt{t})'\big|_{t=2}=\frac{1}{2\sqrt{t}}\big|_{t=2}=\frac{\sqrt{2}}{4}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.