Understanding the algebra behind these limit problems

[Arnoldjavs3]

Homework Statement

$$\lim_{x \to -∞}{\sqrt{x^2 + bx + c} - x}.$$

Homework Equations

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.

 

[member 587159]

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

 

[Arnoldjavs3]

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

Wouldn't that give me b/0?

 

[Mark44]

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.

 

[Arnoldjavs3]

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

 

[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?

 

[Mark44]

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.

 

[Arnoldjavs3]

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.

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)

 

[Mark44]

So basically multiply it by $$\frac{\sqrt{x^2+bx+c}- x}{\sqrt{x^2+bx+c}- x}$$ ?
$$\lim_{x\to -∞}\frac{2x^2+bx+c}{\sqrt{x^2+bx+c}-x}$$

What I said was "you DON'T need to multiply by the conjugate over itself."

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)

Also, the work just above isn't kosher -- you don't plug in $\infty$.

 

[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?

 

[Ray Vickson]

Homework Statement

$$\lim_{x \to -∞}{\sqrt{x^2 + bx + c} - x}.$$

Homework Equations

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:
$$\lim_{x \to -∞}\frac{bx+c}{\sqrt{x^2+bx+c}+x}$$

$$\lim_{x \to -\infty} \left[ \sqrt{x^2+bx+c}-x \right] = \lim_{t \to +\infty} \left[ \sqrt{t^2 - bt +c} + t \right],$$
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 $\sqrt{t^2} = t$.)

 

[Mark44]

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

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.

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?

 

[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.

 

[Arnoldjavs3]

$$\lim_{x \to -\infty} \left[ \sqrt{x^2+bx+c}-x \right] = \lim_{t \to +\infty} \left[ \sqrt{t^2 - bt +c} + t \right],$$
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 $\sqrt{t^2} = t$.)

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

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.

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

 

[Mark44]

$$\lim_{x \to -\infty} \left[ \sqrt{x^2+bx+c}-x \right] = \lim_{t \to +\infty} \left[ \sqrt{t^2 - bt +c} + t \right],$$
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 $\sqrt{t^2} = t$.)

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

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

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.

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

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 $\frac{a + b}c \cdot \frac{a - b}{a - b} = \frac{a^2 - b^2}{c(a - b)}$.