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I do not understand how absolute value effects this, and why is what I have done wrong. Is the way I tackled the problem correct or am I totally wrong?

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I have looked at this post limits, but the definition of absolute value I dont get, I thought |-2|=2, the same would |-x|=x. What am I not getting?

Community
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ALEXANDER
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    It is not necessarily true that ${|a| \over x} = |{a \over x}|$ (if $x<0$ then ${|a| \over x} = -|{a \over x}|$). – copper.hat Jul 09 '14 at 05:37

2 Answers2

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Suppose you didn't know whether $x$ is positive or negative, then how will you analyse the expression $\Large \frac{2x-5}{|3x+2|}$?

You will have to consider two cases:

(1)If $x$ is positive, then $\Large \frac{2x-5}{|3x+2|}=\Large \frac{2x-5}{3x+2}$

(2)If $x$ is negative, then $\Large \frac{2x-5}{|3x+2|}=\Large \frac{2x-5}{-(3x)+2}$

But in your example, we are given that $x\rightarrow-\infty$

Since $x\rightarrow-\infty$, what you are doing is, dividing numerator by negative value, which is $x$ and denominator by positive value, which is $|x|$.

Vikram
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A long way to go about it is:$$\begin{align} \lim_{x\to-\infty}\frac{2x-5}{|3x+2|} &=\lim_{x\to-\infty}\frac{2x}{|3x+2|}-\lim_{x\to-\infty}\frac{5}{|3x+2|} \\ &=2\lim_{x\to-\infty}\frac{x}{|3x+2|}-5\lim_{x\to-\infty}\frac{1}{|3x+2|} \\ &=2\lim_{x\to-\infty}\frac{x}{|3x+2|}-\frac{5}{\lim_{x\to-\infty} |3x+2|}. \end{align}$$

Now, for the first limit, we have an indeterminate form, and therefore we must use L'Hopital's Rule,
$$\begin{align} 2\lim_{x\to-\infty}\frac{x}{|3x+2|}&=2\lim_{x\to-\infty}\frac{1}{\frac{d}{dx}|3x+2|} \\ &=2\lim_{x\to-\infty}\frac{1}{\frac{3(2+3x)}{|2+3x|}} \\ &= 2\lim_{x\to-\infty}\frac{|3x+2|}{3(3x+2)} \\ &= \frac{2}{3}\lim_{x\to-\infty}\frac{|3x+2|}{3x+2}. \end{align}$$

Now, for any quotient of the form $$Q=\frac{|ax+b|}{ax+b} \ \ \ \ a,b\in\mathbb{R^+}$$ we note that $Q>0$ when $x>-b/a$ and $Q<0$ when $x<-b/a$. Now, since the limit is being taken as $x\to-\infty$, then it is clear that $x<-b/a$, or $x<-2/3$ and therefore, $$\frac{2}{3}\lim_{x\to-\infty}\frac{|3x+2|}{3x+2}=-\frac{2}{3}.$$

Also, regarding the other limit, we see that $$\lim_{x\to-\infty}\frac{5}{|3x+2|}=0$$ since it is a quantity over an infinitely larger (although negative) quantity. Therefore, we can see that $$\lim_{x\to-\infty}\frac{2x-5}{|3x+2|}=-\frac{2}{3}.$$

user138999
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