"Determine the Infinite Limit: lim[(x+2)/(x+3)] as x approaches 3 from the right."
My book says it's negative Infinity, and I don't know how to get this. I got 5/6.... What's the correct process to solve this, to "determine the infinite limit"?
Any and all help is greatly appreciated!

- amonoconnor

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- anonymous

whats the equation u have

- amonoconnor

lim\[\lim_{x \rightarrow 3+}[(x+2)/(x+3)]\]

- anonymous

dame idk maybe @cliffordtapp

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## More answers

- thomas5267

The book probably meant x approaches *-3* from the right since 5/6 is the correct limit.

- amonoconnor

5/6 is correct? Thank you so much!

- thomas5267

I believe your book is quite bad since the term infinite limit refers to the limit as x approaches infinity or negative infinity according to my understanding.

- anonymous

yes he right

- thomas5267

Since x is approaching a finite number, it would not be a infinite limit although the value of such limit is finite. That is:
\[
\lim_{x\to\color{red}{\infty}}f(x)\quad\text{Infinite limit}\\
\lim_{x\to\color{red}{-\infty}}f(x)\quad\text{Infinite limit}\\
\lim_{x\to\color{red}{3}}f(x)=\infty \quad\text{Finite limit}\\
\]

- amonoconnor

Wait, maybe I really don't know how to do these, but wouldn't it be -1/0 if evaluated with x approaching -3+?

- thomas5267

EDIT: Not fully awake.
Since x is approaching a finite number, it would not be a infinite limit although the value of such limit is *infinite*. That is:

- thomas5267

One does not simply divide by zero.

- amonoconnor

Right, obviously... but I'm not "dividing", that's how it would work out to, right?

- amonoconnor

\[\lim_{x \rightarrow -3+}(x+2)/\lim_{x \rightarrow -3+}(x+3) = (-3+2)/(-3+3) = (-1/0) ?\]

- thomas5267

Dividing by zero is a meaningless expression in mathematics. If directly evaluating the limit yields \(\dfrac{-1}{0}\) or \(\dfrac{1}{0}\), then you have to find the limit in other ways.

- amonoconnor

Alright. So... what other ways/avenues?

- thomas5267

@freckles I will leave this to you. I have no idea how to explain the evaluation of a limit with infinite as its value without using the delta-epsilon thing.

- freckles

\[\lim_{x \rightarrow -3^+} \frac{x+2}{x+3} \\ \text{ of } -3^+ \text{ means to the right of -3 } \\ \text{ so } x>-3 \\ \text{ now adding 2 on both sides gives } x+2>-3+2=-1 \\ \text{ this means } x+2 \text{ is near } -1 \\ \\ \text{ now adding both sides by 3 we have } x+3>0 \\ \text{ we already know } \lim_{x \rightarrow -3^+}\frac{x+2}{x+3} \text{ is one of the infinities } \\ \text{ and we know the result is } \frac{negative }{positive } \infty \]

- freckles

if we had
\[\lim_{x \rightarrow -3^-}\frac{x+2}{x+3} \\ \text{ the } -3^- \text{ means we are looking to the left of -3 }\]
\[\text{ so } x<-3 \\ \text{ which means } x+3<0 \text{ which says } x+3 \text{ is negative } \\ \text{ now adding 2 on both sides } x+2<-1 \\ \text{ so for this one we have } \frac{negative }{negative } \infty \]

- freckles

and of course neg/pos=neg
and neg/neg=pos

- amonoconnor

I think I understand everything except for where you said that "we already know that (the equation of the limit) is one of the infinities" ...What does that mean? How do you know that?

- freckles

if you have something/0
where the top something isn't 0
then you know the limit is one of the infinities

- amonoconnor

That wouldn't mean that the Limit doesn't exist?
:/

- freckles

infinity is not a number
so yes you could say the limit does not exist but sometimes the instructor or it is also more meaningful to say what kind of non-existence the limit is if you can

- amonoconnor

So it's like a limit can be an integer that a function STOPS at, but reaches on a graph/data set nonetheless, but sometimes, the "STOPPING" point is the arbitrarily unreachable value that the graph just gets infinitely close to, in this case -infinity?

- freckles

the limit being infinity or -negative infinity at x=some finite number
is that you have a vertical asymptote there
I think something people say the graph blows up there if that makes more sense
for example f(x)=1/x looks like:
|dw:1442259490849:dw|
notice at x=0 we have a vertical asymptote
for x>0 the y values are climbing up with non-stop (y values are getting bigger)
for x<0 the y values are falling down with non-stop (y values are getting negative big)
so we say
\[\lim_{x \rightarrow 0^+} \frac{1}{x}= +\infty \text{ or you can just say } \infty \\ \lim_{x \rightarrow 0^-}\frac{1}{x}=-\infty \\ \text{ but yeah it is true in either case you could say } \\ \text{ the limit does not exist } \\ \]
you would definitely say
\[\lim_{x \rightarrow 0}\frac{1}{x} \text{ does not exist }\]

- freckles

for x>0 the y values are climbing up with non-stop (y values are getting bigger)
for x<0 the y values are falling down with non-stop (y values are getting negative big)
(by the way I'm saying this as we approach x=0)

- amonoconnor

|dw:1442259591580:dw|

- freckles

|dw:1442259887019:dw|

- amonoconnor

Is this what it would look like here? I know what an asymptote is, I just don't understand where the infinity is here. Is it that these graphs will, as they approach -3 on the x axis, move closer and closer to negative infinity on the negative y-axis, because -3 is an asymptote?

- amonoconnor

Gotcha. So the "Limit", L, = -infinity, which is a y-value (technically), derived from all of this from finding 0 in the denominator when standard algebraic evaluation was used? Therefore, we understand that it is an asymptote, where x approaches?

- freckles

\[\lim_{x \rightarrow a^+} f(x)=-\infty \text{ means as } x \text{ approaches } x=a \text{ from the right } \\ \text{ the } y \text{ values will keep getting negative large }\]
basically it means you will see the y values decrease with non-stop as you approach -3 from the right

- freckles

I think I will agree to that

- freckles

but sometimes when we have 0 in the denominator this doesn't always mean we have a vertical asymptote

- freckles

it could mean we have a hole if we also have a factor of 0 on top when pluggin in the number that x approaches

- thomas5267

\[
\lim_{x\to \color{red}{c}}f(x) \text{ exists if and only if }\lim_{x\to \color{red}{c^+}}f(x)=\lim_{x\to \color{red}{c^-}}f(x)
\]

- freckles

\[\lim_{x \rightarrow 3} \frac{x-3}{x-3} \\ \text{ for example here if you plug in 3 you get } \frac{0}{0} \\ \text{ but notice } \frac{x-3}{x-3} =1 \text{ when } x \neq 3 \\ \text{ so } \lim_{x \rightarrow 3} \frac{x-3}{x-3} = \lim_{x \rightarrow 3}1 =1 \\ \text{ so here we do not have a vertical asymptote at } x=3 \\ \text{ but we have a hole at } x=3 \\ \text{ for the graph of } f(x)=\frac{x-3}{x-3}\]

- amonoconnor

The original problem specified that it was from the right, but yes, I agree. You have to check both sides if it's general.

- freckles

so if you plug in your number and you get something/zero (where top something isn't 0) then you know you have one of the infinities
if you plug in your number and you get zero/zero you have more work to do. you could still have limit that exists or a limit that doesn't exist for this case...

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