# Finding the limit, multiplication by the conjugate

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.

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

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