Sign of an Infinite Limit Answer

[Jason76]

So if $\displaystyle f(x)$ and $\displaystyle g(x)$ have different signs, then negative infinity, but if the same, then positive. Given $\displaystyle f(x)/g(x)$

Can someone provide an example? How can $\displaystyle g(x)$ be a certain type of sign, when it would always be $\displaystyle 0$, since a number over $\displaystyle 0$ is infinity (when taking a limit).

 

[Deleted member 4993]

So if $\displaystyle f(x)$ and $\displaystyle g(x)$ have different signs, then negative infinity, but if the same, then positive. Given $\displaystyle f(x)/g(x)$

Can someone provide an example? How can $\displaystyle g(x)$ be a certain type of sign, when it would always be $\displaystyle 0$, since a number over $\displaystyle 0$ is infinity (when taking a limit).

I do not quite understand your question - but may be you are looking for something like $\displaystyle \displaystyle \lim{x \to 0} \frac{sin(x)}{x^2}$ where the denominator is always positive.

 

[Jason76]

I do not quite understand your question - but may be you are looking for something like $\displaystyle \displaystyle \lim{x \to 0} \frac{sin(x)}{x^2}$ where the denominator is always positive.

In this case, you would have positive over positive, so that would be $\displaystyle \infty$ as the answer. What would be an example where it comes out as $\displaystyle -\infty$

I think it goes like this:

The sign of the 0 in the denominator is determined by the positive or negative sign connected to the x value (to the right of it) that approaches a number. On the other hand, the sign of the denominator is simply whatever it computes to.

For instance, $\displaystyle \lim x \rightarrow 5-$ would make the denominator negative, when considering $\displaystyle \dfrac{x + 3}{x - 5} = \dfrac{8}{0-}$, this would come out to $\displaystyle -\infty$ because we have two different signs in the denominator and numerator. Is that right ?

Basically, what I'm asking is "How do you determine the sign of infinity, in the case of problems which have an answer of infinity?"

 

[Deleted member 4993]

In this case, you would have positive over positive, so that would be $\displaystyle \infty$ as the answer. What would be an example where it comes out as $\displaystyle -\infty$

I think it goes like this:

The sign of the 0 in the denominator is determined by the positive or negative sign connected to the x value (to the right of it) that approaches a number. On the other hand, the sign of the denominator is simply whatever it computes to.

For instance, $\displaystyle \lim x \rightarrow 5-$ would make the denominator negative, when considering $\displaystyle \dfrac{x + 3}{x - 5} = \dfrac{8}{0-}$, this would come out to $\displaystyle -\infty$ because we have two different signs in the denominator and numerator. Is that right ?

Basically, what I'm asking is "How do you determine the sign of infinity, in the case of problems which have an answer of infinity?"

The sign of "infinity" can flip/flop depending on which way you are approaching limit.

If we approach the limit (x →5⁺) from the right

$\displaystyle \displaystyle \lim_{x \to 5^{+}}\frac{x-3}{x-5} \ = \ +\infty$

If we approach the limit (x →5^-) from the left

$\displaystyle \displaystyle \lim_{x \to 5^{-}}\frac{x-3}{x-5} \ = \ - \infty$

So the answer to your question - how do you determine the sign of infinity - is work out the answer algebraically. Work with the algebra step by step and you'll get there. To quote Dr. Phil

that is all	[SIZE=-2] [/SIZE]
I know on earth, and all I need to know.

 

[Jason76]

The sign of "infinity" can flip/flop depending on which way you are approaching limit.

If we approach the limit (x →5⁺) from the right

$\displaystyle \displaystyle \lim_{x \to 5^{+}}\frac{x-3}{x-5} \ = \ +\infty$

If we approach the limit (x →5^-) from the left

$\displaystyle \displaystyle \lim_{x \to 5^{-}}\frac{x-3}{x-5} \ = \ - \infty$

So the answer to your question - how do you determine the sign of infinity - is work out the answer algebraically. Work with the algebra step by step and you'll get there. To quote Dr. Phil

that is all
I know on earth, and all I need to know.

How about

1. $\displaystyle \displaystyle \lim_{x \to 5^{-}}\frac{3 - x}{x-5} \$

OR

2. $\displaystyle \displaystyle \lim_{x \to 5^{+}}\frac{3 - x}{x-5} \$

Trying to find some pattern to go by.

 

[Deleted member 4993]

How about

1. $\displaystyle \displaystyle \lim_{x \to 5^{-}}\frac{3 - x}{x-5} \$

OR

2. $\displaystyle \displaystyle \lim_{x \to 5^{+}}\frac{3 - x}{x-5} \$

Trying to find some pattern to go by.

Like I said - only pattern is algebra! Why don't you try those two using algebra?

 

[Jason76]

1. $\displaystyle \displaystyle \lim_{x \to 5^{-}}\frac{3 - x}{x-5} \ = \dfrac{-2}{0-} = \infty$

OR

2. $\displaystyle \displaystyle \lim_{x \to 5^{+}}\frac{3 - x}{x-5} \ = \dfrac{-2}{0+} = -\infty$

Correct?

 

[HallsofIvy]

g is NOT 0, it is approaching 0. If it is approaching 0 "from above", g is positive, if it is approaching 0 "from below", g is negative.

 

[Jason76]

g is NOT 0, it is approaching 0. If it is approaching 0 "from above", g is positive, if it is approaching 0 "from below", g is negative.

So is my answer correct? I understand that $\displaystyle 0+$ means positive in denominator (approaching from above), and $\displaystyle 0-$ negative in the denominator (approaching from below).

 

[Deleted member 4993]

1. $\displaystyle \displaystyle \lim_{x \to 5^{-}}\frac{3 - x}{x-5} \ = \dfrac{-2}{0-} = \infty$

OR

2. $\displaystyle \displaystyle \lim_{x \to 5^{+}}\frac{3 - x}{x-5} \ = \dfrac{-2}{0+} = -\infty$

Correct?

Correct ..... see you don't need the crutch of pattern - just need algebra. Pattern changes - algebra does not.