What is my next step? lim[x->3] (x^3 - 9x) / (x^4 - 3x^2 - x + 3)

[Phatfoo]

I've been sitting on this one for ages and can't figure out what to do next. I know what the answer is (9/13), but don't know how to get there after this step.

Any pointers?

 

[pka]

I've been sitting on this one for ages and can't figure out what to do next. I know what the answer is (9/13), but don't know how to get there after this step.

Any pointers?

View attachment 10383

I would factor this as:

\(\displaystyle x^4-3x^3-x+3 =x^3(x-3)-(x-3) \)

. . . . .\(\displaystyle =(x^3-1)(x-3)\)

. . . . .\(\displaystyle =(x-1)(x^2+x+1)(x-3)\)

 

[JeffM]

I've been sitting on this one for ages and can't figure out what to do next. I know what the answer is (9/13), but don't know how to get there after this step.

Any pointers?

View attachment 10383

A different approach from pka's. Forget factoring.

At x = 3, both the numerator and the denominator equal 0. That means each has a factor of (x - 3) by the fundamental theorem of algebra and the zero-product property. Divide both by (x - 3).

\(\displaystyle (x^3 - 9x) \div (x - 3) = x^2 + 3x.\)

\(\displaystyle (x^4 - 3x^3 - x + 3) \div (x - 3) = x^3 - 1.\)

\(\displaystyle \therefore \dfrac{ x^3 - 9x}{x^4 - 3x^3 - x + 3} = \dfrac{x^2 + 3x}{x^3 - 1} \text { if } x \ne 3.\)

\(\displaystyle \therefore x \approx 3 \implies \dfrac{ x^3 - 9x}{x^4 - 3x^3 - x + 3} \approx \\ \dfrac{3^2 + 3 * 3}{3^3 - 1} = \dfrac{9 + 9}{27 - 1} =

\dfrac{18}{26} = \dfrac{9}{13}.\)

To simplify a rational function, you need to factor the numerator and denominator. When you find that a rational function = 0 / 0 at x = a and you want to find the limit at x = a, you immediately know that (x - a) is a factor of both numerator and denominator, and you never need to factor. Just divide that common (x - 3) out.

Follow?

EDIT: To be rigorous:

\(\displaystyle \displaystyle \lim_{x \rightarrow 3} (x^2 + 3x) = 18.\)

\(\displaystyle \displaystyle \lim_{x \rightarrow 3} (x^3 - 1) = 26.\)

There is a theorem:

\(\displaystyle \displaystyle \lim_{x \rightarrow a} f(x) \in \mathbb R \text,\ \lim_{x \rightarrow a} g(x) \in \mathbb R \text { and } \lim_{x \rightarrow a} g(x) \ne 0 \implies\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a} \dfrac{f(x)}{g(x)} = \lim_{x \rightarrow a} f(x) \div \lim_{x \rightarrow a} g(x).\)

 

[stapel]

I've been sitting on this one for ages and can't figure out what to do next. I know what the answer is (9/13), but don't know how to get there after this step.

Any pointers?

View attachment 10383

Try using what you learned back in algebra:

. . . . .\(\displaystyle x^4\, -\, 3x^3\, -\, x\, +\, 3\)

. . . . .\(\displaystyle x^3\, (x\, -\, 3)\, -\, 1\, (x\, -\, 3)\)

. . . . .\(\displaystyle (x\, -\, 3)\, (x^3\, -\, 1)\)

Then, after factoring in pairs, apply the formula you memorized for factoring the difference of cubes.

 

[pka]

I've been sitting on this one for ages and can't figure out what to do next. I know what the answer is (9/13), but don't know how to get there after this step.

Any pointers?

View attachment 10383

To ms staple, thank you for editing my first post. I have used TeX/LaTeX since mid 80's. But I have more trouble here at this site than ever before.

Comment 1: Note that in the title of the post the denomator of the limit is given as \(\displaystyle x^4-3x^{\large{\bf{2}}}-x+3\).
But on the posted work sheet it is \(\displaystyle x^4-3x^{\large{\bf{3}}}-x+3\).
The second one has to be correct give to factor \(\displaystyle (x-3)\) giving \(\displaystyle \dfrac{0}{0}\).
Comment 2: It was suggested that factoring was not necessary. But I suspect whoever set this question expected the question to be about factoring.