Approaching Positive or Negative Infinity?

[Jason76]

The problem below has a limit that DNE. How do we find out whether it approaches positive or negative infinity? Hint.

\(\displaystyle g(x) = x^{2} - 1\), if \(\displaystyle x < 0\)

\(\displaystyle g(x) = 2x + 1\), if \(\displaystyle x\) is greater than or \(\displaystyle = 0\)

Find \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0\)

\(\displaystyle g(x) = (0)^{2} - 1 = -1\) Left hand limit \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0-\)

\(\displaystyle g(x) = 2(0) + 1 = 1\), \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0+\)

Limits don't match so \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0\) DNE.

 

[Jason76]

The limit goes toward \(\displaystyle \infty\), but how do we know that?

 

[Jason76]

How do we know it doesn't have asymptotes at points of discontinuity?

 

[JeffM]

How do we know it doesn't have asymptotes at points of discontinuity?

We know that because the left and right limits are both real numbers.

\(\displaystyle \displaystyle \lim_{x \rightarrow 0^-}(x^2 - 1) = 0 - 1 = - 1.\) The function is approaching a point, not a line.

\(\displaystyle \displaystyle \lim_{x \rightarrow 0^+}(2x + 1) = 2 + 1 = 3.\) Again the function is approaching a point, not a line.

The reason that this function is not continuous at x = 0 is because it has no limit as x approaches 0, and the reason for that is not because the right and left limits do not exist; they do exist, but one of them is not equal to f(0).

Does that make sense?

 

[Jason76]

This is the problem I'm really trying to ask, and Iv'e figured it out.

Assuming that you got a DNE for a limit and it's asymptotic, then in order to know whether it goes to positive or negative infinity, without looking at the graph, you do this:

Look at the quotient of the final working of the equation.

If at the end you got:

\(\displaystyle n\) = some number

\(\displaystyle z\) = some number

\(\displaystyle \lim x \rightarrow z = [\dfrac{n}{0-}] = -\infty\)

\(\displaystyle \lim x \rightarrow z = [\dfrac{-n}{0-}] = \infty\)

\(\displaystyle \lim x \rightarrow z = [\dfrac{n}{0+}] = \infty\)

\(\displaystyle \lim x \rightarrow z = [\dfrac{-n}{0+}] = -\infty\)

 

[JeffM]

This is the problem I'm really trying to ask, and Iv'e figured it out.

Assuming that you got a DNE for a limit and it's asymptotic, then in order to know whether it goes to positive or negative infinity, without looking at the graph, you do this:

Look at the quotient of the final working of the equation.

If at the end you got:

\(\displaystyle n\) = some number

\(\displaystyle z\) = some number

\(\displaystyle \lim x \rightarrow z = [\dfrac{n}{0-}] = -\infty\)

\(\displaystyle \lim x \rightarrow z = [\dfrac{-n}{0-}] = \infty\)

\(\displaystyle \lim x \rightarrow z = [\dfrac{n}{0+}] = \infty\)

\(\displaystyle \lim x \rightarrow z = [\dfrac{-n}{0+}] = -\infty\)

I am not sure that I grasp what you are saying. It seems a bit fuzzy to me.

For example,

\(\displaystyle \displaystyle f(x) = \dfrac{1}{x} \implies \lim_{x \rightarrow 0^-}f(x) = - \infty\ and\ \lim_{x \rightarrow 0^+}f(x) = + \infty.\)

So what can you say about the limit as x approaches zero? Nothing as far as I am concerned. You can't say it is plus infinity or negative infinity. You can only say that it does not exist because the right and left limits are not equal.