1
$\begingroup$

I need to find $$\lim_{x\to 1} \frac{2-\sqrt{3+x}}{x-1}$$

I tried and tried... friends of mine tried as well and we don't know how to get out of:

$$\lim_{x\to 1} \frac{x+1}{(x-1)(2+\sqrt{3+x})}$$

(this is what we get after multiplying by the conjugate of $2 + \sqrt{3+x}$)

How to proceed? Maybe some hints, we really tried to figure it out, it may happen to be simple (probably, actually) but I'm not able to see it. Also, I know the answer is $-\frac{1}{4}$ and when using l'Hôpital's rule I am able to get the correct answer from it.

$\endgroup$
  • $\begingroup$ $(a-b)(a+b)=a^2-b^2$ $\endgroup$ – Alex May 24 '13 at 2:20
  • 1
    $\begingroup$ $(2 - \sqrt{3 + x})(2 + \sqrt{3 + x}) = 2^2 - \left(\sqrt{(3 + x)}\right)^2 = 4 - (3 + x) = 4 - 3 - x = 1 - x = -(x - 1)$ $\endgroup$ – Namaste May 24 '13 at 2:27
2
$\begingroup$

Multiplying by the conjugate does indeed work. You just forgot to carry the negative sign throughout. After multiplying by the conjugate, the correct expression is $\frac{1-x}{(x-1)(2+\sqrt{3+x})}$

$\endgroup$
4
$\begingroup$

You had the right idea: the issue is in your simplification of the numerator:

$$\begin{align} (2 - \sqrt{3 + x})(2 + \sqrt{3 + x}) & = 2^2 - \left(\sqrt{(3 + x)}\right)^2 \\ \\ & = 4 - (3 + x) \\ \\ & = 4 - 3 - x \\ \\ & = 1 - x = -(x - 1) \end{align}$$

That gives you $$\begin{align} \lim_{x \to 1} \frac{-(x - 1)}{(x - 1)(2 + \sqrt{3 + x}} & = \lim_{x\to 1} \frac{-1}{2 + \sqrt{3 + x}} & = -\frac 14 \end{align}$$

$\endgroup$
  • $\begingroup$ Clear as always +1 $\endgroup$ – Amzoti May 24 '13 at 12:04
1
$\begingroup$

Multiply both numerator and denominator by $2+\sqrt{3+x}$, simplify, cancel out. I get $-\frac{1}{4}$

$\endgroup$

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.