Limits Problem

[questions77]

I factored the numerator. Am i supposed to plug in -5 for the top and bottom?

 

[yoscar04]

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

 

[skeeter]

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

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

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

 

[pka]

Look at the GRAPH HERE

 

[Jomo]

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]

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.

 

[HallsofIvy]

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.