I need to find the limit of this function and I'm not sure how to proceed
$$ \lim_{x\to 1}\frac{\sqrt {x+3} -\sqrt x}{1-x} $$
I multiply by the conjugate
$$ \lim_{x\to 1}\frac{\sqrt {x+3} -\sqrt x}{1-x}*\frac{\sqrt {x+3} +\sqrt x}{\sqrt {x+3} +\sqrt x} $$
and I am left with
$$ \frac{ {x+3} – x}{(1-x)(\sqrt {x+3} +\sqrt x)} = \frac{3}{(1-x)(\sqrt {x+3} +\sqrt x)} $$
but after this I am stuck there is nothing to cancel and pluging in 1 will give me $\frac{3}{0}$
Thank you everybody it is much clearer now !
Best Answer
When you plug in $x=1$ in the beginning, you have the limit in the form $\frac{1}0$, which directly implies that the limit doesn't exist.
$$\lim_{x\to1^+}\frac{\sqrt{x+3}-\sqrt{x}}{1-x}=-\infty\quad\mbox{and}\quad\lim_{x\to1^-}\frac{\sqrt{x+3}-\sqrt{x}}{1-x}=\infty$$
Here's a simple graph of the function $f(x)=\frac{\sqrt{x+3}-\sqrt{x}}{1-x}$ drawn on Desmos.