Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks in advance = D

share|improve this question
1  
Could you use $\LaTeX$? –  Pedro Tamaroff Mar 12 '13 at 0:51

2 Answers 2

$$\frac{x+\sqrt x-2}{x-1}=\frac{(\sqrt x-1)(\sqrt x+2)}{(\sqrt x-1)(\sqrt x+1)}=\frac{\sqrt x+2}{\sqrt x+1}\xrightarrow[x\to 1]{} \frac{3}{2}$$

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.