Question

When is it negative infinity and when is it positive infinity?

\[\lim_{x \rightarrow -1^+} \frac{ x+2 }{ x^3-x } = \left[ \frac{ 1 }{ 0^+ } \right] = +\infty \]

\[\lim_{x \rightarrow 1^+} \frac{ x+2 }{ x^3-x } = \left[ \frac{ 3 }{ 0^+ } \right] = -\infty \]

Answer 1

For example in these two cases above, I don't understand what determines the sign.

Answer 2

\[\lim_{x \rightarrow 1^+}\frac{x+2}{x^3-x}\]
x+2 is positive for x values near 1
what determines the sign is the bottom
we have x>1 since we are looking at values to the right of 1
subtracting one on both sides gives
x-1>0
and we also know x is positive for values x near x=1
so we also have x(x-1)>0 holds
we also know x+1>0 for values of x near x=1
so x(x-1)(x+1)>0
so we have a pos/pos which equals a pos

Answer 3

also I think you made a sign type-o in the bottom question

Answer 4

You need to find the sign of the denominator x^3-x
when you approach 1 from the right x^3-x is approaching 0 with positive values
do for example 1.001 you will see that x^3-x approaches 0.002 which is postive

Answer 5

if you approach it from the left for example 0.9, x^3-x is having the values -0.17
negative
so approaching 1 from different side give us different results

Answer 6

i don't if you made a typo or not for the first limit

Answer 7

but the concept is the same at any rate

Answer 8

Thank you both, I have a clearer picture of the whole scheme of things.
(There is no typo btw, one is approaching negative one, the other positive one)

Answer 9

then you should look at your bottom limit again because it is incorrect

Answer 10

the bottom limit cannot be negative as I explained above