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.
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$\begingroup$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 $$\begin{align} \lim_{x \to 1} \frac{-(x - 1)}{(x - 1)(2 + \sqrt{3 + x}} & = \lim_{x\to 1} \frac{-1}{2 + \sqrt{3 + x}} & = -\frac 14 \end{align}$$
$\endgroup$ 1 $\begingroup$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 $\frac{1-x}{(x-1)(2+\sqrt{3+x})}$
$\endgroup$ $\begingroup$Multiply both numerator and denominator by $2+\sqrt{3+x}$, simplify, cancel out. I get $-\frac{1}{4}$
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