I have found the answer to the following problem both graphically and numerically to be -1/2, but I can't seem to prove that this is the answer algebraically. There must be a mistake with my algebra, but I'm not seeing it. Help?
I have found the answer to the following problem both graphically and numerically to be -1/2, but I can't seem to prove that this is the answer algebraically. There must be a mistake with my algebra, but I'm not seeing it. Help?
For x < 0 , $\displaystyle \sqrt{x^2} = -x$. There's one mistake...
I saw a similar thread out on the internets:
calculus - When can we plug in values in a limit? - Mathematics - Stack Exchange
$\displaystyle \lim_{x\to-\infty}(\sqrt{x^2+x+1}+x)$
$\displaystyle =\lim_{x\to-\infty}\frac{(\sqrt{x^2+x+1}+x)(\sqrt{x^2+x+1}-x)}{(\sqrt{x^2+x+1}-x)}$
$\displaystyle =\lim_{x\to-\infty}\frac{x^2+x+1-x^2}{(\sqrt{x^2+x+1}-x)}$
$\displaystyle =\lim_{x\to\infty}\frac{-x+1}{(\sqrt{x^2-x+1}+x)}$
$\displaystyle =\lim_{x\to\infty}\frac{\frac{-x+1}{x}}{\frac{(\sqrt{x^2-x+1}+x)}{x}}$
$\displaystyle =\lim_{x\to\infty}\frac{\frac{-x}{x}+\frac{1}{x}}{\frac{\sqrt{x^2-x+1}}{x}+\frac{x}{x}}$
$\displaystyle =\lim_{x\to\infty}\frac{-1+\frac{1}{x}}{\sqrt{\frac{x^2-x+1}{x^2}}+1}$
$\displaystyle =\lim_{x\to\infty}\frac{-1+\frac{1}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}+1}$
$\displaystyle =\lim_{x\to\infty}\frac{-1+\frac{1}{x}}{\sqrt{1+0+0}+1}$
$\displaystyle =-\frac{1}{2}$
from about your third step ...
$\displaystyle \lim_{x \to -\infty} \frac{x+1}{\sqrt{x^2+x+1} - x}$
divide every term by $\displaystyle \sqrt{x^2} = |x|$
$\displaystyle \lim_{x \to -\infty} \frac{\frac{x}{|x|} + \frac{1}{|x|}}{\sqrt{\frac{x^2+x+1}{x^2}} - \frac{x}{|x|}}$
$\displaystyle \lim_{x \to -\infty} \frac{-1}{\sqrt{1 + \frac{1}{x} +\frac{1}{x^2}} - (-1)} = -\frac{1}{2}$