Problem 2-5: Evaluating Indeterminate Form

Written by

Calculate $\displaystyle\lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}$ .

Solution
$\frac{\sqrt{x+2}-2}{x-2}=\frac{0}{0}$
$\displaystyle\lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}= \displaystyle\lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}\times \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}$

$=\displaystyle\lim_{x\to 2}\frac{x+2-4}{(x-2) \times (\sqrt{x+2}+2)}$
$=\displaystyle\lim_{x\to 2}\frac{x-2}{(x-2) \times (\sqrt{x+2}+2)}$

$=\displaystyle\lim_{x\to 2}\frac{1}{(\sqrt{x+2}+2)}=\frac{1}{4}$

indeterminate form

Comments

2 responses to “Problem 2-5: Evaluating Indeterminate Form”

September 6, 2010

chel

ahmm.. can you solved this one..
sqrt of x-2/x-4 as xapproaches to 4

Reply
1. September 6, 2010
  
  Dr. Yaz Z. Li
  
  Here you go:
  $\displaystyle\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\displaystyle\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4} \times \frac{\sqrt{x}+2}{\sqrt{x}+2}$
  $=\displaystyle\lim_{x\to 4}\frac{\sqrt{x}^2-2^2}{(x-4)(\sqrt{x}+2)}=\displaystyle\lim_{x\to 4}\frac{x-4}{(x-4)(\sqrt{x}+2)}$
  $=\displaystyle\lim_{x\to 4}\frac{1}{(\sqrt{x}+2)}=\frac{1}{(\sqrt{4}+2)}=\frac{1}{4}$
  
  Reply