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.

  • 1
    $\begingroup$ $x+3 \to 6$ 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
  • 1
    $\begingroup$ For best results, please learn to format your posts. Formatting tips here. $\endgroup$ – Em. Mar 8 '16 at 11:36

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*}


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

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

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.