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

#### Lexielai

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
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

#### Lexielai

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

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 $$\displaystyle \sqrt{2}$$

$$\displaystyle \lim_{x \to -\infty} \frac{3x-2}{\sqrt{2x^2+1}} = -\frac{3}{\sqrt{2}}$$

1 person

#### Lexielai

Okay, when you put it that way it makes a lot more sense. Thanks!