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

Letting $$F(x) = \frac{\sqrt{3x} - 3}{\sqrt{2x-4}-\sqrt{2}},$$ we have $$F(x) = \frac{\sqrt{3}(\sqrt{x} - \sqrt{3})}{\sqrt{2}(\sqrt{x-2}-1)}.$$ Multiplying numerator and denominator by $\sqrt{x-2} + 1$,

$$F(x) = \frac{ (3)^{1/2} ((x(x-2)^{1/2})+(x)^{1/2}-(3(x-2)^{1/2})-(3)^{1/2})} {\sqrt{2}(x-3)}.$$

Dividing numerator and denominator by $x$ and substituting $3$ for $x$, I get $\frac{0}{\sqrt{2}} = 0$. Is it correct? My textbook does not have answer, one of the site gives the answer as $\frac{1}{\sqrt{2}}.$.

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you may use L'Hopital rule...result is $\frac{1}{\sqrt 2}$ –  pedja Oct 11 '11 at 16:22
    
Check last expression $$\frac{\sqrt{3}\left( \sqrt{x}-\sqrt{3}\right) }{\sqrt{2}\left( \sqrt{x-2} -1\right) }\neq \frac{\sqrt{3}\left( x\sqrt{x-2}+\sqrt{x}-\left( 3\sqrt{x-2}- \sqrt{3}\right) \right) }{\sqrt{2}\left( x-3\right) }$$ –  Américo Tavares Oct 11 '11 at 16:33
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@Vikram: You've been on the site for four months; it seems that every time you post, someone has to spend the effort to turn it into legible LaTeX. Perhaps you might want to take some time to see what people are doing so that you can try to do some of it yourself in the future? –  Arturo Magidin Oct 11 '11 at 16:33
    
@AméricoTavares: True: but the denominator is correct, and evaluates to $0$ at $3$, there is no way to just "substitute 3 for $x$" in the expression, even if it were completely correct. –  Arturo Magidin Oct 11 '11 at 16:34
    
@ArturoMagidin: That's right. My comment was intended only to indicate that there was a mistake in the algebra too. (Or may be in the TeX conversion). –  Américo Tavares Oct 11 '11 at 16:39

2 Answers 2

up vote 5 down vote accepted

Hint: $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$.

Applying the hint twice, $$ F(x)=\sqrt{\frac32}\frac{\sqrt{x}-\sqrt3}{\sqrt{x-2}-1}=\sqrt{\frac32}\frac{x-3}{\sqrt{x}+\sqrt3}\,\frac{\sqrt{x-2}+1}{(x-2)-1}=\sqrt{\frac32}\frac{\sqrt{x-2}+1}{\sqrt{x}+\sqrt3}, $$ hence $$ \lim\limits_{x\to3}\,F(x)=\sqrt{\frac32}\frac2{2\sqrt3}=\frac1{\sqrt{2}}. $$

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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).

$$F(x) = \frac{\sqrt{3}}{\sqrt{2}}\cdot\left(\frac{\quad\frac{\sqrt{x}-\sqrt{3}}{x-3}\quad}{\quad\frac{\sqrt{x-2} - 1}{x-3}\quad}\right).$$ So $$\lim_{x\to 3}F(x) = \frac{\sqrt{3}}{\sqrt{2}}\frac{\lim\limits_{x\to 3}\frac{\sqrt{x}-\sqrt{3}}{x-3}}{\lim\limits_{x\to 3}\frac{\sqrt{x-2}-1}{x-3}} = \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{g'(3)}{h'(3)},$$ 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 $$\lim_{x\to 3}F(x) = \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{\frac{1}{2\sqrt{3}}}{\frac{1}{2}} = \frac{\quad\frac{1}{2}\quad}{\frac{\sqrt{2}}{2}}=\frac{1}{\sqrt{2}}.$$

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