Csdrhrt

Hey so I'm having a bit of a hard time understanding this one.

$lim_{xto -infty} x+sqrt{x^2-3x}$

1) $x+sqrt{x^2-3x}$ * $(frac{x-sqrt{x^2-3x}}{x-sqrt{x^2-3x}})$

2) $frac{x^2-(x^2-3x)}{x-sqrt{x^2-3x}}$

3) $frac{3x}{x-sqrt{x^2(1-frac{3}{x}})}$

4) $frac{3x}{x-sqrt{x^2}*sqrt{1-frac{3}{x}}}$

5) $frac{3x}{x-x(sqrt{1-frac{3}{x}})}$

6) $frac{3}{1-(sqrt{1-frac{3}{x}})}$

Now I would just take the limit, it would result in $frac{3}{1-1}$ which would be undefined. For some reason, the $x$ in the denominator of step 5 should turn into $-(-x)$ which in turn would be positive and therefore be $frac{3}{1+sqrt{1=frac{3}{x}}}$ which would equal $frac{3}{2}$.

I really don't get it. Apparently the $-infty$ would mean that $sqrt{x^2}$ = $-x$. We didn't even evaluate the limit yet.. how does that turn into $-x$, just because we know the limit is negative does not mean we evaluated it yet..., why not simplify until there is no more simplification to be done, which is what I did in my steps, which would evaluate to undefined?

Would love some help, thanks!

The problem is indeed that, for $x<0$, $sqrt{x^2}=-x$, so when you pull $x^2$ outside the square root it must become $-x$ and not $x$. We can assume $x<0$ because we're doing a limit for $xto-infty$, so restricting the function to an interval of the form $(-infty,a)$, for any $a$, is possible and doesn't affect the limit.

You avoid most problem of this kind if you switch to positive infinity ($x=-y$) or to “positive $0$” ($x=-1/t$). With the latter method, the limit becomes
$$
lim_{xto-infty}bigl(x+sqrt{x^2-3x},bigr)=
lim_{tto0^+}left(-frac{1}{t}+sqrt{frac{1}{t^2}+frac{3}{t}}right)=
lim_{tto0^+}frac{sqrt{1+3t}-1}{t}
$$
which should be much easier (either as a derivative or using the usual technique with the conjugate). Here we can simplify $sqrt{t^2}=t$ because we're working in a right neighborhood of $0$.

In such cases, to avoid confusion with signs, I often suggest, at least as a check, to take $y=-x to infty$ and then

$$lim_{xto -infty} x+sqrt{x^2-3x}=lim_{yto infty} -y+sqrt{y^2+3y}$$

and from here we can proceed as usual.

$a^2-b^2=(a-b)(a+b).$

$y:=-x$ , and consider $lim y rightarrow + infty.$

$sqrt{y^2+3y}-y= dfrac{(y^2+3y)-y^2}{sqrt{y^2+3y}+y}=$

$dfrac{3y}{sqrt{y^2+3y}+y}= dfrac{3y}{y(sqrt{1+3/y}+1)}$

$=dfrac{3}{sqrt{1+3/y}+1}.$

Take the limit.

We are near $-infty$, so $x<0$ and $|x|=-x$.

thus

$$x+sqrt{x^2-3x}=x+sqrt{x^2(1-frac 3x)}$$

$$=x+|x|sqrt{1-frac 3x}$$
$$=x(1-sqrt{1-frac 3x})$$
$$=xfrac{frac 3x}{1+sqrt{1-frac 3x}}$$

the limit is $frac 32$.

Using Taylor's expansion:

When x approches $-infty$:

$x + sqrt{x^2-3x} = x - x(1 - frac{3}{2x}+text{o}(x)) = x - x + frac{3}{2}+text{o}(1) rightarrow frac{3}{2}$.

So, yeah

$
begin{align}
lim_{xto-infty}x + sqrt{x^2-3x} =frac{3}{2}.
end{align}
$