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Evaluate : $$\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$$

I've tried some basic algebraic manipulation to get it into a form where I can apply L'Hopital's Rule, but it's still going to be indeterminate form.

This is what I've done so far

\begin{align} \lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x}) &= \lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})\left(\frac{x-\sqrt{x^2 + 2x}}{x-\sqrt{x^2 + 2x}}\right)\\ \\ &= \lim_{x \to \ -\infty} \left(\frac{x^2 - (x^2 + 2x)}{x-\sqrt{x^2 + 2x}}\right)\\ \\ &= \lim_{x \to \ -\infty} \left(\frac{-2x}{x-\sqrt{x^2 + 2x}}\right)\\ \\ \end{align}

And that's as far as I've gotten. I've tried applying L'Hopitals Rule, but it still results in an indeterminate form.

Plugging it into WolframAlpha shows that the correct answer is $-1$

Any suggestions on what to do next?

Martin Sleziak
  • 56,060
Perturbative
  • 13,656

4 Answers4

10

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

haqnatural
  • 22,026
0

HINT:

Set $-1/x=h\implies h\to0^+, h>0$

$x^2+2x=\dfrac{1-2h}{h^2}\implies\sqrt{x^2+2x}=+\dfrac{\sqrt{1-2h}}h$

Now rationalize the numerator to get

$$\dfrac{\sqrt{1-2h}-1}h=\dfrac{1-2h-1}{h(\sqrt{1-2h}+1)}$$

lab bhattacharjee
  • 279,016
0

This is based on @Battani's answer but with a more in-depth explanation

\begin{align} \lim _{ x\to -\infty } \left( \frac { -2x }{ x-\sqrt { x^{ 2 }+2x } } \right) &= \lim _{ x\rightarrow -\infty }\left( \frac { -2x }{ x-\sqrt { { x }^{ 2 }\left( 1+\frac { 2 }{ x } \right) } } \right) \\ \\ &\text{Now because $\sqrt{x^2}$ = $|x|$} \\ \\ &= \lim _{ x\rightarrow -\infty } \frac { -2x }{ x-\left| x \right| \sqrt { 1+\frac { 2 }{ x } } } \\ \\ \text{Recall that } & \ |x| = \begin{cases} x &\text{ if } \ \ x \geq 0\\ - x &\text{ if } \ \ x < 0\\ \end{cases} \\ \\ &\text{$x$ is approaching $-\infty$} \\ \\ &\therefore \ \ \ \ |x| = -x \\ \\ &= \lim _{ x\rightarrow -\infty } \frac { -2x }{ x- (-x)\sqrt { 1+\frac { 2 }{ x } } } \\ \\ &= \lim _{ x\rightarrow -\infty } \frac { -2x }{ x + x\sqrt { 1+\frac { 2 }{ x } } } \\ \\ &= \lim _{ x\rightarrow -\infty } \frac { -2x }{ x\left( 1+\sqrt { 1+\frac { 2 }{ x } } \right) } \\ \\ &= \lim _{ x\rightarrow -\infty } \frac { -2 }{ 1+\sqrt { 1+\frac { 2 }{ x } } } \\\\ &= -1 \end{align}

Perturbative
  • 13,656
0

This solution is probably flawed, but I will post this anyway. Note that $$\lim_{x \to \ -\infty} \left(x + \sqrt{x^2 + 2x}\right)$$ $$=\lim_{x \to \ -\infty} \left(x + \sqrt{(x+1)^2-1}\right).$$ The $-1$ in the radical is basically negligible compared to the $(x+1)^2$, so let us ignore that. Doing this gives $$\lim_{x \to \ -\infty} \left(x + \sqrt{(x+1)^2}\right)$$ $$=\lim_{x \to \ -\infty} \left(x \pm (x+1)\right)$$ $-(x+1)$ is positive, while $x+1$ is negative for sufficiently low $x$. Square roots are always nonegative for real $x$, so we should add $-(x+1)$. Doing so gives $$\lim_{x \to \ -\infty} \left(x - (x+1)\right)$$ $$=\lim_{x \to \ -\infty} -1$$ $$=\boxed{-1}.$$

Jeffrey H.
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