# Indeterminate Limit (Finding the Remainder to a Root)

So i was working on this:

$$\lim\limits_{x\to1} \frac{x + \sqrt{x} - 2}{x - 1}$$

and I thought to simpify my top by multiplying by a conjugate, taking everything other than the $x$ to be the $b$ from $a+b$ so that my conjugate looked like $x - \sqrt{x} + 2$.

The multiplication, if correct, led me to $x^2 - x + 4\sqrt{x} - 4$, which i was happy to note had $1$ as a root (which corresponded with my denominator and would allow for me to cancel off the [x-1]'s.

However, I'm having trouble finding the chunk that multiplies $(x-1)$ to give the $x^2 - x + 4\sqrt{x} - 4$, and this chunk's limit will define the nominator's limit, which, divided by 2 (from the conjugate left alone at the denominator after I cancel) should be my overall limit, i think = P

I'm pretty new to calculus so I may have made some mistakes, but if I am, I'd still like to know how I would have gotten that other root, since Ruffini's isn't working for me.

Could you use $\LaTeX$? – Pedro Tamaroff Mar 12 '13 at 0:51
$$\frac{x+\sqrt x-2}{x-1}=\frac{(\sqrt x-1)(\sqrt x+2)}{(\sqrt x-1)(\sqrt x+1)}=\frac{\sqrt x+2}{\sqrt x+1}\xrightarrow[x\to 1]{} \frac{3}{2}$$
Hint: I think you will find things much easier if you let $x=t^2$. This makes no real mathematical difference, but will send you in the right direction. The $\sqrt{x}$ was causing unnecessary confusion. After the substitution, the work will take seconds only.