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I'm trying to solve the following limit.


$$\lim \limits_{x \to 2}{\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}}$$


Unfortunately, I haven't made any headway.

I tried multiplying by the conjugates of both the numerator and denominator (in 2 different attempts, of course), but ended up with a monster of an expression either time.

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marked as duplicate by Workaholic, LutzL, Shailesh, SBareS, Willie Wong Nov 18 '16 at 14:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Mattos -- I did; and I got a monster of an expression. :) $\endgroup$ – Fine Man Nov 18 '16 at 5:07
  • $\begingroup$ Can you use L'Hospital's rule? If so its easy $\endgroup$ – R_D Nov 18 '16 at 5:07
  • $\begingroup$ @R_D -- No, not yet. I'm going through a calculus textbook, but so far I've only learnt about limits (pretty much as advanced as The Squeeze Theorem). $\endgroup$ – Fine Man Nov 18 '16 at 5:08
  • $\begingroup$ @SirJony Sorry, I probably should have read your question in its entirety. $\endgroup$ – mattos Nov 18 '16 at 5:08
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    $\begingroup$ Duplicate: $\displaystyle\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}$. (Found using Approach0.xyz) $\endgroup$ – Workaholic Nov 18 '16 at 9:28
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Let $y = 3 - x$. Then we have \begin{align} \lim_{x \to 2} \frac{\sqrt{6-x} - 2}{\sqrt{3 - x} - 1} &= \lim_{y \to 1} \frac{\sqrt{y + 3} - 2}{\sqrt{y} - 1}\\ &= \lim_{y \to 1} \frac{(y + 3 - 2^2)(\sqrt{y} + 1)}{(y - 1)(\sqrt{y + 3} + 2)}\\ &= \lim_{y \to 1} \frac{(y - 1)(\sqrt{y} + 1)}{(y-1)(\sqrt{y + 3} + 2)}\\ &= \lim_{y \to 1} \frac{\sqrt{y} + 1}{\sqrt{y + 3} + 2} = \frac{1 + 1}{2 + 2} = \frac{1}{2} \end{align}

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  • $\begingroup$ @SirJony I multiplied by the conjugate. Sorry I skipped a bit of simplication. Similar to $\frac{a - b}{c - d} = \frac{(a - b)(a+b)(c+d)}{(c-d)(a+b)(c+d)} = \frac{(a^2 - b^2)(c+d)}{(c^2 - d^2)(a+b)}$. $\endgroup$ – Jeevan Devaranjan Nov 18 '16 at 5:20
  • $\begingroup$ Oh, I see. Also, how'd you figure this out!? I would've never thought of this. $\endgroup$ – Fine Man Nov 18 '16 at 5:26
  • $\begingroup$ @SirJony When solving problems try to see if a substitution can make the problem simpler. Radicals are normally a pain, so you want to reduce the amount of terms inside them. Recognizing when to use a substitution comes with experience. Soon enough you will be able to solve problems like these in no time! $\endgroup$ – Jeevan Devaranjan Nov 18 '16 at 6:00
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HINT:

$$\lim_{x\to2}\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x\to2}\dfrac{6-x-2^2}{3-x-1^2}\cdot\lim_{x\to2}\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$$

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You can apply L'Hospital Rule because after putting limit, we get, $$\frac{0}{0}$$ Applying L'Hospital Rule, $$\lim_{x\to2}\frac{\frac{-1}{2\sqrt{6-x}}}{\frac{-1}{2\sqrt{3-x}}}$$ After simplifying, you get, $$\lim_{x\to2}\frac{\sqrt{3-x}}{\sqrt{6-x}}$$ Applying limits, $$\frac{1}{2}$$

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I like to let my variables approach zero. So I would do this like this:

$\begin{array}\\ \lim \limits_{x \to 2}\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} &=\lim \limits_{y \to 0}\dfrac{\sqrt{6-(y+2)}-2}{\sqrt{3-(y+2)}-1} \qquad(x = y+2)\\ &=\lim \limits_{y \to 0}\dfrac{\sqrt{4-y}-2}{\sqrt{1-y}-1}\\ &=\lim \limits_{y \to 0}\dfrac{2\sqrt{1-y/4}-2}{\sqrt{1-y}-1}\\ &=\lim \limits_{y \to 0}\dfrac{2(1-y/8+O(y^2))-2}{(1-y/2+O(y^2))-1}\\ &=\lim \limits_{y \to 0}\dfrac{-y/4+O(y^2)}{-y/2+O(y^2)}\\ &=\lim \limits_{y \to 0}\dfrac{-1/4+O(y)}{-1/2+O(y)}\\ &=\dfrac12\\ \end{array} $

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