# Limits Problem

#### yoscar04

##### Full Member
You should show us what have you tried so far.
Try to factorize the numerator. If you get stuck, write back.

#### skeeter

##### Elite Member
$$\displaystyle \lim_{x \to -5} \dfrac{(x+5)(x-3)}{|x+5|}$$

you need to look at both one sided limits ...

$$\displaystyle \lim_{x \to -5^-} \dfrac{(x+5)(x-3)}{|x+5|}$$ and $$\displaystyle \lim_{x \to -5^+} \dfrac{(x+5)(x-3)}{|x+5|}$$

• yoscar04 and Jomo

#### pka

##### Elite Member
• BraveHeart258

#### Jomo

##### Elite Member
If x=-5 makes both the numerator and denominator equal to 0, then one factor for the numerator and denominator will be x+5.

Also check both sided limits and cancel carefully.

What does |x+5| equal without the absolute value bars?

#### HallsofIvy

##### Elite Member
I factored the numerator and plugged in -5 and the denominator ends up with 0
Yes, it does. Now what about the numerator? $$\displaystyle (-5)^2+ 2(-5)- 15= 25- 10- 15= 0$$ also so both numerator and denominator are 0!
View attachment 21600

Does the limit not exist for this?
[/QUOTE]
Think about $$\displaystyle \lim_{x\to 0} \frac{x}{x}$$. The denominator goes to 0 (as does the numerator) but, for x not 0, this is precisely 1. It's limit exists and is 1.

You say you factored the numerator and, I presume, you got (x+ 5)(x- 3). $$\displaystyle \frac{x^3+ 2x- 15}{|x+ 5|}= \frac{(x+ 5)(x- 3)}{|x+ 5|}$$.

For x> -5, $$\displaystyle |x+ 5|= x+ 5$$ so this is $$\displaystyle \frac{(x+ 5)(x- 3)}{x+ 5}= x- 3$$. The limit of that, for x going to -5, is -5- 3= -8.

For x< -5. $$\displaystyle |x+ 5|= -(x+ 5)$$ so this is $$\displaystyle \frac{(x+ 5)(x-3)}{-(x+ 5)}= -(x- 3)$$. The limit of that, for x going to -5, is $$\displaystyle -(-5- 8)= 8$$.

The limit does not exist but because the two one-sided limits are not the same, not because the denominator is 0. If the problem had been $$\displaystyle \lim_{x\to -5}\frac{x^3+ 2x- 15}{x+ 5}$$, without the absolute value, the limit would be -8 even though the denominator still goes to 0.

• BraveHeart258 and Jomo

#### HallsofIvy

##### Elite Member
Yes, it does. Now what about the numerator? $$\displaystyle (-5)^2+ 2(-5)- 15= 25- 10- 15= 0$$ also so both numerator and denominator are 0!
View attachment 21600

Does the limit not exist for this?
Think about $$\displaystyle \lim_{x\to 0} \frac{x}{x}$$. The denominator goes to 0 (as does the numerator) but, for x not 0, this is precisely 1. It's limit exists and is 1.

You say you factored the numerator and, I presume, you got (x+ 5)(x- 3). $$\displaystyle \frac{x^3+ 2x- 15}{|x+ 5|}= \frac{(x+ 5)(x- 3)}{|x+ 5|}$$.

For x> -5, $$\displaystyle |x+ 5|= x+ 5$$ so this is $$\displaystyle \frac{(x+ 5)(x- 3)}{x+ 5}= x- 3$$.[/quote]
I think one should specifically say that "$$\displaystyle \frac{(x+ 5)(x- 3)}{x+ 5}= x- 3$$" is only true for x not equal to 5. Of course, since the "limit as x goes to 5" does not depend upon the value at x= 5, the limit is still correct.

The limit of that, for x going to -5, is -5- 3= -8.

For x< -5. $$\displaystyle |x+ 5|= -(x+ 5)$$ so this is $$\displaystyle \frac{(x+ 5)(x-3)}{-(x+ 5)}= -(x- 3)$$. The limit of that, for x going to -5, is $$\displaystyle -(-5- 8)= 8$$.

The limit does not exist but because the two one-sided limits are not the same, not because the denominator is 0. If the problem had been $$\displaystyle \lim_{x\to -5}\frac{x^3+ 2x- 15}{x+ 5}$$, without the absolute value, the limit would be -8 even though the denominator still goes to 0.