Why Do You Divide Bottom Denominator of Limit by -X When it Approach -Infinity?

Jun 2014
23
0
United States
Hi everyone!

I was doing a homework assignment involving limits today, and something came up that didn't make any sense at all to me. In the example for finding the lim (x-> -Infinity) (3x-2)/sqrt(2x^2+1), the examples says to divide both the numerator and denominator by x. It does this in a peculiar way for the denominator though. After dividing, the resultant equation becomes:

[ lim (x -> -Infinity) (3 - 2/x) ]/[lim (x -> -Infinity) sqrt(2x^2 + 1)/-sqrt(x^2)

Now, I know the sqrt(x^2) is just x, but the inconsistency I see is that they divided the bottom by -sqrt(x^2) instead of the positive. Furthermore, the top remains divided by positive x. This means that they're dividing the equation by x on the numerator and -x on the denominator, making it an equation divided by -1. I was under the impression that you're allowed to do whatever with 1, but not -1.

I read something about it being because x->-Infinity, which is why they use -x, but that still doesn't make sense. The numerator still needs to be divided by the same number as the denominator, so why don't they divide by -x and go with x instead?

It really frustrates me to see something that feels inconsistent like this. If anyone could shed light on why they do this, it would be much appreciated!
 

topsquark

Forum Staff
Jan 2006
11,579
3,455
Wellsville, NY
Hi everyone!

I was doing a homework assignment involving limits today, and something came up that didn't make any sense at all to me. In the example for finding the lim (x-> -Infinity) (3x-2)/sqrt(2x^2+1), the examples says to divide both the numerator and denominator by x. It does this in a peculiar way for the denominator though. After dividing, the resultant equation becomes:

[ lim (x -> -Infinity) (3 - 2/x) ]/[lim (x -> -Infinity) sqrt(2x^2 + 1)/-sqrt(x^2)

Now, I know the sqrt(x^2) is just x, but the inconsistency I see is that they divided the bottom by -sqrt(x^2) instead of the positive. Furthermore, the top remains divided by positive x. This means that they're dividing the equation by x on the numerator and -x on the denominator, making it an equation divided by -1. I was under the impression that you're allowed to do whatever with 1, but not -1.

I read something about it being because x->-Infinity, which is why they use -x, but that still doesn't make sense. The numerator still needs to be divided by the same number as the denominator, so why don't they divide by -x and go with x instead?

It really frustrates me to see something that feels inconsistent like this. If anyone could shed light on why they do this, it would be much appreciated!
Is this part of a larger proof? As far as I can determine you are correct.

Did they follow the error through and still get the correct answer?

-Dan
 
Jun 2014
23
0
United States
I'm pulling this example from my textbook, which goes on to say that [lim (x -> -Infinity) (3 - 2/x)]/[lim (x -> -Infinity) sqrt(2x^2 + 1)/-sqrt(x^2) = lim (3-2/x)/-sqrt(2+1/x^2) = 3-0/-sqrt(2+0) = -3/sqrt(2)

My teacher did something similar with another problem, where she divided the bottom by -sqrt(x^2), or -x. This appears again in a homework problem, worked out at this website: CalcChat | Calculus 9e | Easy Access Study Guide (the book is Calculus 9e, Chapter 3, Section 5, Exercise #27).

The answer is apparently correct, but I don't understand the logic behind why it is. Calculus sure likes to mess with me :c
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
\(\displaystyle \lim_{x \to -\infty} \frac{3x-2}{\sqrt{2x^2+1}}\)

note ... \(\displaystyle \sqrt{x^2} = |x|\)

\(\displaystyle \lim_{x \to -\infty} \frac{\frac{3x}{|x|}-\frac{2}{|x|}}{\frac{\sqrt{2x^2+1}}{\sqrt{x^2}}}\)

\(\displaystyle \lim_{x \to -\infty} \frac{\frac{3x}{|x|}-\frac{2}{|x|}}{\sqrt{\frac{2x^2}{x^2}+\frac{1}{x^2}}}\)

\(\displaystyle \lim_{x \to -\infty} \frac{\frac{3x}{|x|}-\frac{2}{|x|}}{\sqrt{2+\frac{1}{x^2}}}\)

since \(\displaystyle x \to -\infty\), the first fraction in the numerator divides out to -3 and the second fraction approaches 0

as \(\displaystyle x \to -\infty\) , the denominator approaches \(\displaystyle \sqrt{2}\)


\(\displaystyle \lim_{x \to -\infty} \frac{3x-2}{\sqrt{2x^2+1}} = -\frac{3}{\sqrt{2}}\)
 
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Jun 2014
23
0
United States
Okay, when you put it that way it makes a lot more sense. Thanks!