Question

Does a limit that gives X/0 ( x being any number except than 0 ) will be infinity ( negative or positive depending on x) ? Or can it also not exist at all?

Answer 1

I think it doesn't exist, but I guess you can interpretate as positive infinite to x>0 and negative infinite to x<0

Answer 2

Or it depends upon your definition of limit?

Answer 3

I had this problem here lim x--> 2 on the left hand side (3x+4) ^2 / x ^2-x-2 gave me - infinity. Denominator factors out to x+1 & x-2. So it cannot be cancelled out in someway. The answer key gives me a negative. thing is I don't understand why some problems do that :/

Answer 4

It's -infinite to \[x \rightarrow2^-\] and +infinite to -infinite to \[x \rightarrow2^+\]
http://tutorial.math.lamar.edu/Classes/CalcI/InfiniteLimits.aspx

Answer 5

thanks :) ill read it ill give you a best response :)

Answer 6

I had a mistake while writting the response.
Is -infinite when
\[x \rightarrow 2^-\]
And +infinite when:
\[x \rightarrow 2^+\]

Ignore the "-infinite to" after the "+infinite to"

Answer 7

\[\lim\limits_{x\rightarrow 2}\quad\frac{ (3x+4) ^2 }{ x ^2-x-2}\]\[=\lim\limits_{x\rightarrow 2}\quad\frac{ (3x+4) ^2 }{ (x+1)(x-2)}\]

Answer 8

it all depends on values of left limit and right limit.
If left limit = right limit = + infinity
then limit EXIST and equals +infinity

If left limit = right limit = - infinity
then limit EXIST and equals -infinity

if left limit NOT equal to right limit, then limit DOES NOT EXIST.
no matter what the values of left and right limit are.
In your case, u had (x-2) in denominator.
for left limit x->2- means x is very very near to -2 but less. (like -1.99999)
means x-2 will be negative.
Hence, the answer to left limit is -infinity.
Does all this make sense ?

Answer 9

Sorry for the late answer I was working haha. and Yes it does make more sense, as x approaches 2, the number will be increasingly bigger but because the limit would be over 0 it's an asymptote if I understood well enough :)