1
$\begingroup$

$$\lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(\frac{1}{\sqrt{4 + \epsilon}} - \frac{1}{2}\right)$$

I know the limit is $-\frac{1}{16}$, but I just can't figure out how comes... Can anyone help me with a hint?

$\endgroup$
0
$\begingroup$

HINT : Yours will be $$\frac 1\epsilon\times\frac{2-\sqrt{4+\epsilon}}{2\sqrt{4+\epsilon}}=\frac 1\epsilon \times\frac{4-(4+\epsilon)}{2\sqrt{4+\epsilon}\times \left(2+\sqrt{4+\epsilon}\right)}=\frac{-1}{2\sqrt{4+\epsilon}\times \left(2+\sqrt{4+\epsilon}\right)}$$

$\endgroup$
  • $\begingroup$ You are welcome! $\endgroup$ – mathlove Jan 6 '14 at 17:32

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.