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The question is to evaluate the given limit :

$$\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}$$

When I'm trying to evaluate this I'm getting $1$ as my answer but the answer given in the text is $0$. I'm wondering how can this limit be equal to $0$.

I'm using $x = 3 + h$ (assuming $h$ is approaching zero) and now converting this limit in the form of $h$. Lim (h tends to zero) and substituting $x = 3 + h$.

In this way numerator equals $h$ and now after rationalizing the denominator I will get $2h$ and the numerator will also become $2h$ both of them will cancel each other and the limit will be equal to $1$.

I don't know where I'm going wrong. Please edit if you can for proper reading because I don't know how to edit this in mathematical format.

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    $\begingroup$ $x+3 \to 6$ as $x \to 3$. Did you mean $x-3$ in the numerator? $\endgroup$ – Henry Mar 8 '16 at 11:15
  • $\begingroup$ @Henry yeah the edited question looks the same. How can this limit be equal to 0. I'm solving it by right approach. $\endgroup$ – Saksham Mar 8 '16 at 11:19
  • $\begingroup$ It must be $x-3$ in numerator. $\endgroup$ – Claude Leibovici Mar 8 '16 at 11:24
  • $\begingroup$ @ClaudeLeibovici yeah changes made. $\endgroup$ – Saksham Mar 8 '16 at 11:26
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    $\begingroup$ For best results, please learn to format your posts. Formatting tips here. $\endgroup$ – Em. Mar 8 '16 at 11:36
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We also have \begin{align*} \lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}&=\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}\cdot\frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}}\\ &=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(x-2)-(4-x)}\\ &=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2x-6}\\ &=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}\\ &=\lim_{x \to 3}\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\\ &=\frac{\sqrt{3-2}+\sqrt{4-3}}{2}\\ &=\frac{\sqrt{1}+\sqrt{1}}{2}\\ &=\frac{2}{2}\\ &=1. \end{align*}

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Your working is correct. The answer is one. This is probably one of those cases where the textbook has a typo in it.

$$\lim_{x\to3}\frac{x+3}{\sqrt{x-2}-\sqrt{4-x}}=\lim_{h\to0}\frac{h}{\sqrt{1+h}-\sqrt{1-h}}$$ $$=\lim_{h\to0}\frac{h(\sqrt{1+h}+\sqrt{1-h})}{(\sqrt{1+h}-\sqrt{1-h})(\sqrt{1+h}+\sqrt{1-h})}$$ $$=\lim_{h\to0}\frac{h(\sqrt{1+h}+\sqrt{1-h})}{2h}$$ $$=\lim_{h\to0}\frac{(\sqrt{1+h}+\sqrt{1-h})}{2}$$ $$=\frac{1+1}{2}$$ $$=1$$

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  • $\begingroup$ That's exactly how I was solving it. Apart from the question I just want to know how can I write my question in mathematical format like others have written their answers. $\endgroup$ – Saksham Mar 8 '16 at 11:38
  • $\begingroup$ You mean how to type it here at Math.SE? $\endgroup$ – Ian Miller Mar 8 '16 at 11:40
  • $\begingroup$ yeah I don't know how to write that way but now someone has shared a link the comments. $\endgroup$ – Saksham Mar 8 '16 at 11:53
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One thing that's wrong is the + sign in the numerator. Most likely it should be a - sign as suggested by @Henry.

If I multiply the numerator and denominator by $\sqrt{x-2}+\sqrt{4-x}$ I get a difference of squares in the new denominator, whose value cancels the $x-3$ factor. Then I do in fact get 1. 0 and 1 are next to each other on a QWERTY board and somebody's finger slipped.

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  • $\begingroup$ Yeah it's - in the numerator and I have edited it. $\endgroup$ – Saksham Mar 8 '16 at 11:30

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