#### questions77

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Look at the GRAPH HERE

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Yes, it does. Now what about the numerator? \(\displaystyle (-5)^2+ 2(-5)- 15= 25- 10- 15= 0\) also soI factored the numerator and plugged in -5 and the denominator ends up with 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.

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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.Yes, it does. Now what about the numerator? \(\displaystyle (-5)^2+ 2(-5)- 15= 25- 10- 15= 0\) also sobothnumerator and denominator are 0!

View attachment 21600

Does the limit not exist for this?

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.