Problem 2-5: Evaluating Indeterminate Form

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}

Problem 2-5

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    Comments

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

    1. chel Avatar
      chel

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

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