Question

Find the limit as x approaches negative infinity of (x+sqrt(x^2 + x +1)) ?

Sorry to sound needy, but HEEEEEEEEELP.

Answer 1

\[\lim_{x\to-\infty}\left(x+\sqrt{x^2+x+1}\right)\]
\[\begin{align*}x+\sqrt{x^2+x+1}\cdot\frac{x-\sqrt{x^2+x+1}}{x-\sqrt{x^2+x+1}}&=\frac{x^2-\left(x^2+x+1\right)}{x-\sqrt{x^2+x+1}}\\
&=\frac{-x-1}{x-\sqrt{x^2+x+1}}\\
&=\frac{-x-1}{x-\sqrt{x^2}\sqrt{1+\frac{1}{x}+\frac{1}{x^2} }}\\
&=\frac{-x-1}{x-|x|\sqrt{1+\frac{1}{x}+\frac{1}{x^2} }}
\end{align*}\]
Since \(x\to-\infty\), you have \(|x|=-x\):
\[\begin{align*}\frac{-x-1}{x-(-x)\sqrt{1+\frac{1}{x}+\frac{1}{x^2} }}&=\frac{-x-1}{x+x\sqrt{1+\frac{1}{x}+\frac{1}{x^2} }}\\
&=\frac{-1-\frac{1}{x}}{1+\sqrt{1+\frac{1}{x}+\frac{1}{x^2} }}
\end{align*}\]
Taking the limit should be easy now.

Answer 2

You again! Thanks. :D

Answer 3

You're welcome : )