Question

"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!

Answer 1

whats the equation u have

Answer 2

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

Answer 3

dame idk maybe @cliffordtapp

Answer 4

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

Answer 5

5/6 is correct? Thank you so much!

Answer 6

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.

Answer 7

yes he right

Answer 8

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

Answer 9

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

Answer 10

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:

Answer 11

One does not simply divide by zero.

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

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

Answer 16

@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.

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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?

Answer 25

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:

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
\[\lim_{x \rightarrow 0}\frac{1}{x} \text{ does not exist }\]