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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

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  • $\begingroup$ Show us the multiplication you did, and the complex equation you ended up with, we'll see if we can straighten you out. $\endgroup$ Sep 14, 2012 at 3:20
  • $\begingroup$ i ended up with this puu.sh/14T7n $\endgroup$
    – Maria
    Sep 14, 2012 at 3:24
  • $\begingroup$ 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! $\endgroup$ Sep 14, 2012 at 3:29
  • $\begingroup$ @user1561559: Shouldn't the second factor in the numerator be $\sqrt{3-x}+1$? $\endgroup$ Sep 14, 2012 at 3:36
  • $\begingroup$ i tried to conjugate again, but it returned back to its original form! :( am i doing something wrong? $\endgroup$
    – Maria
    Sep 14, 2012 at 3:38

4 Answers 4

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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.

$$\begin{align*} \lim_{x \to 2} \;\frac{\sqrt{6-x} - 2}{\sqrt{3-x} - 1} \;&=\; \lim_{x \to 2} \;\frac{\bigl((6-x) - 4\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl((3-x) - 1\bigr)} &\qquad\qquad\tag{multiply by conjugates} \\[2ex]&=\; \lim_{x \to 2} \;\frac{\bigl(2-x\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl(2 - x\bigr)} &\qquad\qquad\tag{simplify} \\[2ex]&=\; \lim_{x \to 2} \;\frac{\sqrt{3-x} + 1}{\sqrt{6-x} + 2} &\qquad\qquad\tag{simplify more} \\[2ex]&=\; \frac{\lim_{x \to 2} \bigl(\sqrt{3-x} + 1\bigr)}{\lim_{x \to 2} \bigl(\sqrt{6-x} + 2\bigr)} \tag{as both limits exist $\ldots$} \\[2ex]&=\; \frac{1 + 1}{2 + 2} \;=\; \frac{1}{2}. \tag{$\ldots$ by substitution} \end{align*}$$

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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.

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  • $\begingroup$ Isn't your recommendation to use the derivatives just L'L'Hospital's rule? $\endgroup$ Sep 14, 2012 at 3:27
  • $\begingroup$ @Ross: No, it is using the definition of the derivative, and the properties of limits of products and reciprocals. $\endgroup$ Sep 14, 2012 at 3:28
  • $\begingroup$ we haven't learnt derivatives yet :S $\endgroup$
    – Maria
    Sep 14, 2012 at 3:30
  • $\begingroup$ @user1561559: Oops. $\endgroup$ Sep 14, 2012 at 3:30
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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 $(1)$: $$ \begin{align} \lim_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} &=\lim_{x\to2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\\ &=\frac12\tag{2} \end{align} $$

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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.

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    $\begingroup$ 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. $\endgroup$ Sep 14, 2012 at 3:43
  • $\begingroup$ alright i got it! thanks everyone!! $\endgroup$
    – Maria
    Sep 14, 2012 at 4:07

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