# How do you find the limit of x-sqrt(x^2+3x) as x approaches negative infinity?

Jun 26, 2016

$- \infty$

#### Explanation:

${\lim}_{x \setminus \to - \infty} \setminus x - \sqrt{{x}^{2} + 3 x}$

let $z = - x$

$\setminus \implies {\lim}_{z \setminus \to \infty} \setminus - z - \sqrt{{z}^{2} - 3 z}$

$\setminus \implies {\lim}_{z \setminus \to \infty} \setminus - z + \textcolor{red}{{\lim}_{z \setminus \to \infty} - \sqrt{{z}^{2} - 3 z}}$

the black LHS clearly goes to $- \infty$

for the RHS in red, we note that $\sqrt{{z}^{2} - 3 z} \ge 0 \setminus \forall z \in \left[3 , \infty\right)$

therefore, the RHS clearly also goes to $- \infty$

so the limit is $- \infty$