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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}$2011-10-11
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    Check last expression $$\frac{\sqrt{3}\left( \sqrt{x}-\sqrt{3}\right) }{\sqrt{2}\left( \sqrt{x-2} -1\right) }\neq \frac{\sqrt{3}\left( x\sqrt{x-2}+\sqrt{x}-\left( 3\sqrt{x-2}- \sqrt{3}\right) \right) }{\sqrt{2}\left( x-3\right) }$$2011-10-11
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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?2011-10-11
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    @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.2011-10-11
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    @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).2011-10-11
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    @AméricoTavares:It may have been the LaTeXing; it's a bit hard to do; I'll check... Okay, I just cut-n-pasted from the original post, only putting brackets around the exponents; looks like it's the same thing.2011-10-11
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    As a note, you can use "view source" to see other people's TeX, even for posts where you don't have the 'edit' privilege.2011-10-11

2 Answers 2

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