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.