Krdytkyu

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?

$$lim _{ xto -infty } left( frac { -2x }{ x-sqrt { x^{ 2 }+2x } } right) =lim _{ xrightarrow -infty }{ left( frac { -2x }{ x-sqrt { { x }^{ 2 }left( 1+frac { 2 }{ x } right) } } right) =lim _{ xrightarrow -infty }{ frac { -2x }{ x-left| x right| sqrt { 1+frac { 2 }{ x } } } = } } \ =lim _{ xrightarrow -infty }{ frac { -2x }{ x+xsqrt { 1+frac { 2 }{ x } } } = } lim _{ xrightarrow -infty }{ frac { -2x }{ xleft( 1+sqrt { 1+frac { 2 }{ x } } right) } = } -1$$

HINT:

Set $-1/x=himplies hto0^+, h>0$

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

Now rationalize the numerator to get

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

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

begin{align}
lim _{ xto -infty } left( frac { -2x }{ x-sqrt { x^{ 2 }+2x } } right) &= lim _{ xrightarrow -infty }left( frac { -2x }{ x-sqrt { { x }^{ 2 }left( 1+frac { 2 }{ x } right) } } right) \ \
&text{Now because $sqrt{x^2}$ = $|x|$} \ \
&= lim _{ xrightarrow -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 _{ xrightarrow -infty } frac { -2x }{ x- (-x)sqrt { 1+frac { 2 }{ x } } } \ \
&= lim _{ xrightarrow -infty } frac { -2x }{ x + xsqrt { 1+frac { 2 }{ x } } } \ \
&= lim _{ xrightarrow -infty } frac { -2x }{ xleft( 1+sqrt { 1+frac { 2 }{ x } } right) } \ \
&= lim _{ xrightarrow -infty } frac { -2 }{ 1+sqrt { 1+frac { 2 }{ x } } } \\
&= -1
end{align}

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
$$lim_{x to -infty} left(x - (x+1)right)$$
$$=lim_{x to -infty} -1$$
$$=boxed{-1}.$$