# Limit of quotients with square roots: $\lim_{x\to2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$

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

• Show us the multiplication you did, and the complex equation you ended up with, we'll see if we can straighten you out. Sep 14, 2012 at 3:20
• i ended up with this puu.sh/14T7n Sep 14, 2012 at 3:24
• Good - I take it you got there by multiplying top and bottom by the conjugate of the denominator. Now do the same with the conjugate of the numerator. Have faith! Sep 14, 2012 at 3:29
• @user1561559: Shouldn't the second factor in the numerator be $\sqrt{3-x}+1$? Sep 14, 2012 at 3:36
• i tried to conjugate again, but it returned back to its original form! :( am i doing something wrong? Sep 14, 2012 at 3:38

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.

• Isn't your recommendation to use the derivatives just L'L'Hospital's rule? Sep 14, 2012 at 3:27
• @Ross: No, it is using the definition of the derivative, and the properties of limits of products and reciprocals. Sep 14, 2012 at 3:28
• we haven't learnt derivatives yet :S Sep 14, 2012 at 3:30
• @user1561559: Oops. Sep 14, 2012 at 3:30

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}

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.

• Leaving $\sqrt{3-x}+1$ in the numerator as is, and also using the conjugate of $\sqrt{6-x}-2$, allows for a more straightforward solution. Sep 14, 2012 at 3:43
• alright i got it! thanks everyone!! Sep 14, 2012 at 4:07