Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 20

Determine $\displaystyle \lim_{x \to -\infty} (x + \sqrt{x^2 + 2x})$

We apply completing the square in the equation under the square root.


$
\begin{equation}
\begin{aligned}

x^2 + 2x =& 0
\\
\\
x^2 + 2x + 1 =& 1
\\
\\
(x + 1)^2 =& 1
\\
\\
(x + 1)^2 - 1 =& 0
\\
\\
\lim_{x \to - \infty} x^2 + 2x =& \lim_{x \to - \infty} (x + 1)^2 - 1 = \lim_{x \to - \infty} (x + 1)^2
\qquad \text{We can ignore -1 because its too small compared to $(x + 1)^2$ as $x$ approaches infinity.}

\end{aligned}
\end{equation}
$


So, we can rewrite the limit


$
\begin{equation}
\begin{aligned}

\lim_{x \to - \infty} x + \sqrt{x^2 + 2x} =& \lim_{x \to - \infty} x + \sqrt{(x + 1)^2}
\\
\\
=& \lim_{x \to - \infty} x + | x + 1 |
\qquad \text{when $x$ approaches $- \infty$, it makes $(x + 1)$ negative, so $| x + 1 | = - (x + 1)$}
\\
\\
=& \lim_{x \to - \infty} x - (x + 1)
\\
\\
=& \lim_{x \to - \infty} x - x - 1
\\
\\
=& \lim_{x \to - \infty} -1
\\
\\
=& - 1

\end{aligned}
\end{equation}
$