The rule for evaluating limits of rational functions by dividing the coefficients of highest powers

Question

I have a Limit problem as below:

Connor claims: " $\lim_{x\to \infty} \left(\frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}\right) = 3$ because my high school calculus teacher told us the limit of ratio of polynomials is always the quotient of the coefficients of the highest power terms"

If correct, show in detail how to use algebra and the limit theorems to evaluate this limit and get the same answer

If wrong,

1, use algebra and limit theorems to correctly evaluate the limit and

2, write a paragraph that explain why Connor shouldn't expect the rule he remember from high school to work in this particular problem

I am able to correctly evaluate the limit ( lim = 0) but since my first language is not English, I don't understand what Connor claimed and how to explain it. Can anyone help? Thanks in advance

Answer 1

The rule applies when the highest power in the numerator and the highest power in the denominator are the same. But here the highest power in the numerator is $2$ and the highest power in the denominator is $3$. So the rule doesn't apply, and the correct limit is $0$ as you said.

(Of course there is another "high school rule" for this situation, it's just a different rule.)

Answer 2

If you are able to evaluate limits like this, you should remember/know that the following is true: $$ \lim_{x\to \infty} \frac{ax^{\pmb n} + \sum_{i < n} a_i x^i}{bx^{\pmb n} + \sum_{i < n} b_i x^i} = \frac{a}b $$ that is what happend, if the polynomials in numerator and denominator have the same degree. Connor wrongfully states ("remembers") that $$ \lim_{x\to \infty} \frac{ax^{\pmb m} + \sum_{i < m} a_i x^i}{bx^{\pmb n} + \sum_{i < n} b_i x^i} = \frac{a}b \tag{$\color{red}{\rm wrong}$} $$ holds for all $n$ and $m$.

Answer 3

$$\frac{16x^2}{2x^3} > \frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}>\frac{6x^2}{3x^3} $$

for all large enough $x$. Now, by the squeeze theorem,

$$0 = \lim_{x\to\infty}\frac{16}{2x} = \lim_{x\to\infty}\frac{16x^2}{2x^3} \geq \lim_{x\to\infty}\frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}\geq \lim_{x\to\infty}\frac{6x^2}{3x^3} = \lim_{x\to\infty}\frac{6}{3x}= 0$$

Hence the limit is zero. This is a general rule: If the degree of the denominator is higher than the degree of the numerator, the limit is zero. If you don't want to use this rule, use above derivation.

Answer 4

Theorem: If $f,g: \mathbb{R} \to \mathbb{R}$ such that $f(x) \to a$ for some $a \in \mathbb{R}$ and $g(x) \to b$ for some $b \neq 0$ as $x \to \infty$, then $f(x)/g(x) \to a/b$ as $x \to \infty$.

Applying the theorem, we see that $$ \frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1} = \frac{\frac{6}{x} + \frac{7}{x^{2}} + \frac{3}{x^{3}}}{2 + \frac{1}{x} - \frac{2}{x^{2}} - \frac{1}{x^{3}}} \to 0 $$ as $x$ grows indefinitely.

Answer 5

$$\lim_{x\to \infty} \left(\frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}\right)=$$

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The leading term in the denominator of $\frac{6x^2 + 7x +3}{2x^3 + x^2 -2x -1}$ is $x^3$. Divide the numerator and denominator by this:

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$$\lim_{x\to \infty} \left(\frac{\frac{6}{x}+\frac{7}{x^2}+\frac{3}{x^3}}{2+\frac{1}{x}-\frac{2}{x^2}-\frac{1}{x^3}}\right)=$$

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Te expressions $\frac{3}{x^3},\frac{7}{x^2},\frac{6}{x},-\frac{1}{x^3},-\frac{2}{x^3},\frac{1}{x}$ all tend to zero as $x$ approaches $\infty$:

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$$\frac{0+0+0}{2+0-0-0}=\frac{0}{2}=0$$