L Rule Finding Limits #2

[Silvanoshei]

I'm I setting this up right?

\(\displaystyle \lim \limits_{x\rightarrow 2^{+}} \left[\frac{8}{x^{2}-4}-\frac{2}{x-2}\right] ==> 0-0\)

\(\displaystyle \frac{8(x-2)-2(x^{2}-4)}{(x^{2}-4)(x-2)} ==> Factored ==> (x^{3}-2x^{2}-4x+8)\)

\(\displaystyle \frac{8-4}{3x^{2}-4x-4} ==> \frac{4}{-8} ==> -\frac{1}{2}?\)

 

[JeffM]

I'm I setting this up right?

\(\displaystyle \lim \limits_{x\rightarrow 2^{+}} \left[\frac{8}{x^{2}-4}-\frac{2}{x-2}\right] ==> 0-0\) Wrong.

\(\displaystyle \frac{8(x-2)-2(x^{2}-4)}{(x^{2}-4)(x-2)} ==> Factored ==> (x^{3}-2x^{2}-4x+8)\) No.

\(\displaystyle \frac{8(x - 2) - 2(x^2 - 4)}{(x^2 - 4)(x - 2)} = \frac{8x - 16 - 2x^2 + 8}{(x + 2)(x - 2)(x - 2)} =\)

\(\displaystyle \frac{-2x^2 + 8x - 8}{(x + 2)(x - 2)^2} = \frac{-2(x^2 - 4x + 4)}{(x + 2)(x - 2)^2} = -\frac{2(x - 2)^2}{(x + 2)(x - 2)^2} = -\frac{2}{x + 2}.\)

\(\displaystyle \frac{8-4}{3x^{2}-4x-4} ==> \frac{4}{-8} ==> -\frac{1}{2}?\)

Where do these numbers come from? \(\displaystyle \ 3 * 2^2 - 4 * 2 - 4 = 3 * 4 - 8 - 4 = 12 - 12 = 0 \ne - 8.\)

You could reduce the risk of error by factoring earlier in the game.

\(\displaystyle x \ne 2\ and\ x \ne - 2 \implies \dfrac{8}{x^2 - 4} - \dfrac{2}{x - 2} = \dfrac{8}{(x + 2)(x - 2)} - \dfrac{2}{x - 2} =\)

\(\displaystyle \dfrac{8}{(x + 2)(x - 2)} - \dfrac{2(x + 2)}{(x + 2)(x - 2)} = \dfrac{8 - 2x - 4}{(x + 2(x - 2)} =\)

\(\displaystyle \dfrac{-2x + 4}{(x + 2)(x - 2)} = \dfrac{-2(x - 2)}{(x + 2)(x - 2)} =\)

\(\displaystyle - \dfrac{2}{x + 2}.\)

Same answer, but simpler fractions.

 

[Silvanoshei]

\(\displaystyle \lim \limits_{x\rightarrow 2^{+}} \left[\frac{8}{x^{2}-4}-\frac{2}{x-2}\right] ==> 0-0\)

When you plug the limit value into the equation right away, you will get 0-0, or rather \(\displaystyle \infty - \infty\) which is something you should always do to make sure what you're dealing with.

\(\displaystyle \frac{8(x-2)-2(x^{2}-4)}{(x^{2}-4)(x-2)}\)

Isn't this how you combine fractions by multiplying the denominators together and cross multiplying the numerators minus each other into 1 fraction? I know you used a simpler method which I like better, but, this isn't necessarily wrong?

 

[JeffM]

\(\displaystyle \lim \limits_{x\rightarrow 2^{+}} \left[\frac{8}{x^{2}-4}-\frac{2}{x-2}\right] ==> 0-0\)

When you plug the values into the equation right away, you will get 0-0, which is something you should always do to make sure what you're dealing with.

\(\displaystyle \frac{8(x-2)-2(x^{2}-4)}{(x^{2}-4)(x-2)}\)

Isn't this how you integrate by multiplying the denominators together and cross multiplying the numerators minus each other into 1 fraction? I know you used a simpler method which I like better, but, this isn't necessarily wrong?

\(\displaystyle \displaystyle \lim_{x \rightarrow a}f(x) = b \in \mathbb R\ and\ \lim_{x \rightarrow a}g(x) = c \in \mathbb R\ and\ h(x) = f(x) \pm g(x) \implies \lim_{x \rightarrow a}h(x) = b \pm c.\)

The above is a law of limits, but it does not apply to what you have.

\(\displaystyle \displaystyle \lim_{x \rightarrow 2^+}\dfrac{8}{x^2 - 4} = \infty \notin \mathbb R.\) And \(\displaystyle \displaystyle \lim_{x \rightarrow 2^+}\dfrac{4}{x - 2} = \infty \notin \mathbb R.\)

What you are doing when you manipulate the expression into a form that is equivalent at points where the relevant functions exist is to determine what the function equals close to those points. Do you understand? The function does not exist at those points, but whereever it exists it is equivalent to the other function, WHICH CAN BE EVALUATED at those points. I have to leave so if you do not get this, someone else will have to follow up.

My "No" was not to say the method was wrong, but to point out that you made an error when applying it.