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Aliana Alvarez

Answered question

2022-04-12

Find the limit of a function

lim_{x \to 3} \frac{\sqrt{3 x} - 3}{\sqrt{2 x - 4} - \sqrt{2}}

Raina Blackburn

Beginner2022-04-13Added 10 answers

Your error is that after rationalizing the denominator, you cannot just plug in 3 for x, because the denominator still evaluates to 0 (not

\sqrt{2}

, as you seem to think).

F (x) = \frac{\sqrt{3}}{\sqrt{2}} \cdot (\frac{\frac{\sqrt{x} - \sqrt{3}}{x - 3}}{\frac{\sqrt{x - 2} - 1}{x - 3}})

lim_{x \to 3} F (x) = \frac{\sqrt{3}}{\sqrt{2}} \frac{lim_{x \to 3} \frac{\sqrt{x} - \sqrt{3}}{x - 3}}{lim_{x \to 3} \frac{\sqrt{x - 2} - 1}{x - 3}} = \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{g^{'} (3)}{h^{'} (3)}

where

g (x) = \sqrt{x}

and

h (x) = \sqrt{x - 2}

Since

g^{'} (x) = \frac{1}{2 \sqrt{x}}, g^{'} (3) = \frac{1}{2 \sqrt{3}}

. Since

h^{'} (x) - \frac{1}{2 \sqrt{x - 2}}

, then

h^{'} (3) = \frac{1}{2}

. So

lim_{x \to 3} F (x) = \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\frac{1}{2 \sqrt{3}}}{\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2}}

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