# What is the limit of  (x +sqrt((x^2)+(3x))) as x goes to infinity?

Oct 6, 2015

${\lim}_{x \rightarrow \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = \infty$

#### Explanation:

$\left(x + \sqrt{{x}^{2} + 3 x}\right) = x + \sqrt{{x}^{2} \left(1 + \frac{3}{x}\right)}$ for $x \ne 0$

 = x +sqrt(x^2)sqrt(1+3/x)) for $x \ne 0$

 = x +absx sqrt(1+3/x)) for $x \ne 0$

Now, as $x \rightarrow \infty$, we have

$\left\mid x \right\mid \rightarrow \infty$ and

$\sqrt{1 + \frac{3}{x}} \rightarrow \sqrt{1 + 0} = 1$

So,

${\lim}_{x \rightarrow \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = {\lim}_{x \rightarrow \infty} \left(x + \left\mid x \right\mid \sqrt{1 + \frac{3}{x}}\right) = \infty$

Bonus

As $x$ decreases without bound, the limit is quite different.

As $x \rightarrow - \infty$, we get indeterminate form: $- \infty + \infty$.

We think of $\left(x + \sqrt{{x}^{2} + 3 x}\right)$ as a ratio (over $1$) and eliminate the square root from the numerator.

$\left(x + \sqrt{{x}^{2} + 3 x}\right) \cdot \frac{x - \sqrt{{x}^{2} + 3 x}}{x - \sqrt{{x}^{2} + 3 x}} = \frac{- 3 x}{x - \sqrt{{x}^{2} + 3 x}}$

$= \frac{- 3 x}{x - \left\mid x \right\mid \sqrt{1 + \frac{3}{x}}}$.

For $x < 0$, we have $\left\mid x \right\mid = - x$, so we have:

${\lim}_{x \rightarrow - \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = {\lim}_{x \rightarrow - \infty} \frac{- 3 x}{x - \left\mid x \right\mid \sqrt{1 + \frac{3}{x}}}$

$= {\lim}_{x \rightarrow - \infty} \frac{- 3 x}{x + x \sqrt{1 + \frac{3}{x}}}$

$= {\lim}_{x \rightarrow - \infty} \frac{- 3}{1 + \sqrt{1 + \frac{3}{x}}}$

$= - \frac{3}{2}$.

Oct 6, 2015

${\lim}_{x \to + \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = + \infty$
${\lim}_{x \to - \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) = - \frac{3}{2}$

#### Explanation:

$\sqrt{{x}^{2} + 3 x} \ge 0$ for all $x \ge 0$

So

${\lim}_{x \to + \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right) \ge {\lim}_{x \to + \infty} x = + \infty$

Much more interesting is the case ${\lim}_{x \to - \infty} \left(x + \sqrt{{x}^{2} + 3 x}\right)$

First notice that ${\lim}_{t \to \infty} \left(t - \sqrt{{t}^{2} - C}\right) = 0$ for any constant $C > 0$

To see this: Given $\epsilon > 0$

${\left(t - \epsilon\right)}^{2} = {t}^{2} - 2 t \epsilon + {\epsilon}^{2}$

So if $t > \frac{C + {\epsilon}^{2}}{2 \epsilon}$, then

${\left(t - \epsilon\right)}^{2} = {t}^{2} - 2 t \epsilon + {\epsilon}^{2} < {t}^{2} - \left(C + {\epsilon}^{2}\right) + {\epsilon}^{2} = {t}^{2} - C$

So

$t - \epsilon < \sqrt{{t}^{2} - C} < \sqrt{{t}^{2}} = t$

Let $t = - \left(x + \frac{3}{2}\right)$ and $C = \frac{9}{4}$

Then:

$0 = {\lim}_{t \to \infty} \left(\sqrt{{t}^{2} - C} - t\right)$

$= {\lim}_{x \to - \infty} \left(\sqrt{{\left(x + \frac{3}{2}\right)}^{2} - \frac{9}{4}} + x + \frac{3}{2}\right)$

$= {\lim}_{x \to - \infty} \left(\sqrt{{x}^{2} + 3 x + \frac{9}{4} - \frac{9}{4}} + x + \frac{3}{2}\right)$

$= {\lim}_{x \to - \infty} \left(\sqrt{{x}^{2} + 3 x} + x + \frac{3}{2}\right)$

$= {\lim}_{x \to - \infty} \left(\sqrt{{x}^{2} + 3 x} + x\right) + \frac{3}{2}$

Hence:

${\lim}_{x \to - \infty} \left(\sqrt{{x}^{2} + 3 x} + x\right) = - \frac{3}{2}$