You can't use L'Hospital's rule :S
$\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}$
I've tried to multiply by conjugates but ended up with a so complex equation, please help, anyone? :S
You can't use L'Hospital's rule :S
$\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}$
I've tried to multiply by conjugates but ended up with a so complex equation, please help, anyone? :S
You are right to try conjugates: $\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}=\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}{{\sqrt{3-x}+1}\over {\sqrt{3-x}+1}}=\lim_{x\to2}\frac{\sqrt{(6-x)(3-x)}+\sqrt{6-x}-2\sqrt{3-x}-2}{2-x}$ which is still $\frac 00$, but we can define $u=2-x$ and try a Taylor series
$\lim_{x\to2}\frac{\sqrt{(6-x)(3-x)}+\sqrt{6-x}-2\sqrt{3-x}-2}{2-x}=\lim_{u \to 0} \frac{\sqrt{(4+u)(1+u)}+\sqrt{4+u}-2\sqrt{1+u}-2}{u}=\lim_{u \to 0}\sqrt{1+\frac 5u+4}+\sqrt{\frac4{u^2}+\frac 1u}-\sqrt{\frac 4{u^2}+\frac 4u}-\frac 2u$
Now pull out the terms in $\frac 1u$, which should cancel and you will be left with something finite.
The key is to multiply by the conjugates of both radicals, in order to eliminate what makes the problem annoying: the two radicals whose limits both go to zero. And then you see what cancels out.
$\begin{align*} \lim_{x \to 2} \;\frac{\sqrt{6-x} - 2}{\sqrt{3-x} - 1} \;&=\; \lim_{x \to 2} \;\frac{\bigl((6-x) - 4\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl((3-x) - 1\bigr)} &\qquad\qquad\tag{multiply by conjugates} \\[2ex]&=\; \lim_{x \to 2} \;\frac{\bigl(2-x\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl(2 - x\bigr)} &\qquad\qquad\tag{simplify} \\[2ex]&=\; \lim_{x \to 2} \;\frac{\sqrt{3-x} + 1}{\sqrt{6-x} + 2} &\qquad\qquad\tag{simplify more} \\[2ex]&=\; \frac{\lim_{x \to 2} \bigl(\sqrt{3-x} + 1\bigr)}{\lim_{x \to 2} \bigl(\sqrt{6-x} + 2\bigr)} \tag{as both limits exist $\ldots$} \\[2ex]&=\; \frac{1 + 1}{2 + 2} \;=\; \frac{1}{2}. \tag{$\ldots$ by substitution} \end{align*}$
Multiplying by conjugates works. I suggest not getting discouraged by the complexity, because it works out in the end.
An alternative is to rewrite it as
$ \left(\frac{\sqrt{6-x}-2}{x-2}\right) \cdot \frac{1}{\left(\dfrac{\sqrt{3-x}-1}{x-2}\right)}$
and notice that each parenthesized expression has the form $\dfrac{f(x)-f(2)}{x-2}$ (for two different $f$s). Then you can find the limit by evaluating the derivatives, i.e., limits of difference quotients, and using the properties of limits of products and reciprocals.
This is not any different than some of the other solutions, just hopefully a bit easier to follow: $ \begin{align} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} &=\frac{\color{#C00000}{\sqrt{6-x}-2}}{\color{#00A000}{\sqrt{3-x}-1}} \frac{\color{#C00000}{\sqrt{6-x}+2}}{\color{#C00000}{\sqrt{6-x}+2}} \frac{\color{#00A000}{\sqrt{3-x}+1}}{\color{#00A000}{\sqrt{3-x}+1}}\\ &=\color{#C00000}{\frac{(6-x)-4}{\sqrt{6-x}+2}} \color{#00A000}{\frac{\sqrt{3-x}+1}{(3-x)-1}}\\ &=\frac{\color{#00A000}{\sqrt{3-x}+1}}{\color{#C00000}{\sqrt{6-x}+2}} \frac{\color{#C00000}{2-x}}{\color{#00A000}{2-x}}\\ &=\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\tag{1} \end{align} $ Take $\lim\limits_{x\to2}$ of $(1)$: $ \begin{align} \lim_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} &=\lim_{x\to2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\\ &=\frac12\tag{2} \end{align} $