Problem 2-5: Evaluating Indeterminate Form

Posted byYaz 2 Comments on 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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2 Comments

  1. Reply

    1. Here you go:
      $latex \displaystyle\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\displaystyle\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4} \times \frac{\sqrt{x}+2}{\sqrt{x}+2}$
      $latex =\displaystyle\lim_{x\to 4}\frac{\sqrt{x}^2-2^2}{(x-4)(\sqrt{x}+2)}=\displaystyle\lim_{x\to 4}\frac{x-4}{(x-4)(\sqrt{x}+2)}$
      $latex =\displaystyle\lim_{x\to 4}\frac{1}{(\sqrt{x}+2)}=\frac{1}{(\sqrt{4}+2)}=\frac{1}{4}$

      Reply

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