Homework Help: Understanding the algebra behind these limit problems

1. Nov 3, 2016

Arnoldjavs3

1. The problem statement, all variables and given/known data
$$\lim_{x \to -∞}{\sqrt{x^2 + bx + c} - x}.$$

2. Relevant equations

3. The attempt at a solution
So in problem 1, once I got to a point where I am to divide by the highest power in the denominator(x) I get something like:
$$\lim_{x \to -∞}\frac{bx+c}{\sqrt{x^2+bx+c}+x}$$

Now what I want to clarify for myself is if the following is correct:
$$\lim_{x \to -∞}\frac{bx+c}{\sqrt{x^2}\sqrt{1+b/x+c/x^2}+x}$$
$$\lim_{x \to -∞}\frac{b+c/x}{(|x|/x)\sqrt{1+b/x+c/x^2}+1)}$$
$$\lim_{x \to -∞}\frac{b+0}{(x)/(x)\sqrt{1+0+0}+1}$$
and you get $$b/2$$

So, when you have $$\frac{|x|}{(x)}$$ how would I go about dividing it in this situation? I know that if we are looking at a left sided limit of say 0, |x| = -(x) and vice versa for the right sided limit.

But this limit is at negative infinity, so how would it work in this situation? Please let me know if I'm wrong in my algebra approach here.

Last edited: Nov 3, 2016
2. Nov 3, 2016

Math_QED

$\sqrt{x^2}= -x$ when $x \rightarrow -\infty$

3. Nov 3, 2016

Arnoldjavs3

Wouldn't that give me b/0?

4. Nov 3, 2016

Staff: Mentor

Try multiplying by $\sqrt{x^2 - x} - x$ over itself (i.e., multiplying by 1).

Edit: Original problem was changed, so this advice no longer applies.

Last edited: Nov 3, 2016
5. Nov 3, 2016

Arnoldjavs3

Jesus I had just realized I wrote down the wrong problem. Going to change it

6. Nov 3, 2016

Arnoldjavs3

Okay... so at this point is where I'm confused if I'm wrong/right:

$$\frac{b+c/x}{\sqrt{x^2}\sqrt{1+b/x+c/x^2} + x}$$

$$\frac{b+c/x}{{|x|/x}\sqrt{1+b/x+c/x^2} + x}$$

$$\frac{b+0}{{-(x)/x}\sqrt{1+0+0} + x}$$

and then I get b/0 which is not the correct answer. Where did I go wrong here?

7. Nov 3, 2016

Staff: Mentor

Is this still the problem?

$$\lim_{x \to -∞}{\sqrt{x^2 + bx + c} - x}.$$
If so, you can evaluate it almost directly -- you don't need to multiply by the conjugate over itself, which was my earlier hint.

8. Nov 3, 2016

Arnoldjavs3

So basically multiply it by $$\frac{\sqrt{x^2+bx+c}- x}{\sqrt{x^2+bx+c}- x}$$ ?
$$\lim_{x\to -∞}\frac{2x^2+bx+c}{\sqrt{x^2+bx+c}-x}$$
Then I divided by the highest power in the denominator(x) and evaluated it at -∞ to get this:

$$\lim_{x \to -∞}\frac{2(-∞)+b}{\sqrt{1 +0 + 0} - 1}.$$ Is what I got, which ended up being -∞/0. Where am I messing up? The answer is b/2(which I coincidentally got in the OP but it was done incorrectly)

9. Nov 3, 2016

Staff: Mentor

What I said was "you DON'T need to multiply by the conjugate over itself."
Also, the work just above isn't kosher -- you don't plug in $\infty$.

10. Nov 3, 2016

Arnoldjavs3

$$\frac{\sqrt{x^2+bx+c}- x}{\sqrt{x^2+bx+c}- x}$$ Is not the conjugate

I thought you had meant to multiply both the numerator and the denominator by itself(because this evaluates to 1). Could you further elaborate on what you meant?

11. Nov 3, 2016

Ray Vickson

$$\lim_{x \to -\infty} \left[ \sqrt{x^2+bx+c}-x \right] = \lim_{t \to +\infty} \left[ \sqrt{t^2 - bt +c} + t \right],$$
by taking $x = -t$. Note that $\sqrt{t^2 - b t + c} = \sqrt{t^2} \sqrt{1-z} = t \sqrt{1-z}$, where $z = (bt-c)/t^2 \to 0$ as $t \to \infty$. (Note that for $t > 0$ we have $\sqrt{t^2} = t$.)

12. Nov 3, 2016

Staff: Mentor

Right, it's not the conjugate over itself, but what's the point of doing this? My comment about multiplying by the conjugate over itself (which is 1) applied to the original problem, which you changed.

13. Nov 4, 2016

Arnoldjavs3

My prof stated that this question was supposed to be the limit at infinity, not negative infinity.

My approach was correct in the original post however I put $$|x|/x$$ = $$(x)/x$$ at negative infinity which was my mistake.

14. Nov 4, 2016

Arnoldjavs3

Is there a thereom/name for what you're doing here? I've never seen this before

What were you suggesting with the other problem? Just for reference

15. Nov 4, 2016

Staff: Mentor

AFAIK, this doesn't rise to the level of needing a theorem to justify it. All that's going on is substitution.

If you are working with a limit in the form of $\frac{a + b}c$, and both numerator and denominator evaluate to 0 (or both are infinite), it can sometimes be helpful to multiply by 1 in the form of the conjugate over itself. I.e., work with $\frac{a + b}c \cdot \frac{a - b}{a - b} = \frac{a^2 - b^2}{c(a - b)}$.