Evaluating $\lim_{n\to\infty}\frac{n^3+6n^2}{2n^2+5n+6}$

Question

I was copying my teacher slides and I noticed something weird. If you look at Key Flex answer, it shows why I used zero initially(Look at edit logs). I have noticed the problem in my question, so I will post another question in a new post explaining the problem. So the question is this:

$$\lim\limits_{n\to\infty}\dfrac{n^3+6n^2}{2n^2+5n+6} \tag{1}$$

Steps on how teacher solved it:

$$\lim\limits_{n\to\infty}\dfrac{n^3+6n^2}{2n^2+5n+6} \tag{2}$$

$$\lim\limits_{n\to\infty}\dfrac{n^2(n+6)}{n^2\left(2+\dfrac{5}{n}+\dfrac{6}{n^2}\right)}\tag{3}$$

$$\lim\limits_{n\to\infty}\dfrac{\infty+6}{2+\dfrac{5}{\infty}+\dfrac{6}{\infty^2}} =\infty \tag{4}$$

Shouldn't it be: $$\lim\limits_{n\to\infty}\dfrac{\infty+6}{2+\dfrac{5}{\infty}+\dfrac{6}{\infty^2}} =\dfrac{6}{2}=3 \tag{5}$$

Answer 1

When taking infinite limits of "polynomial-over-polynomial" expressions, the polynomial with the bigger degree wins. (In the case of a tie, the coefficients give the answer.)

In the simplest case, we have $$\lim_{n\to\infty}\frac{n^p}{n^q} \quad=\quad \lim_{n\to\infty}\begin{cases} n^{p-q} & \text{if}\;p>q \\[4pt] 1 & \text{if}\;p=q \\ \dfrac{1}{n^{q-p}} & \text{if}\;p<q \end{cases} \quad=\quad \begin{cases} \infty \\[6pt] 1 \\[6pt] 0 \end{cases}$$ which reflects the fundamental limiting nature of infinity. (The (positive) power of larger and larger numbers gets larger and larger; and the reciprocal of the (positive) power of larger and larger numbers gets smaller and smaller. (And, of course, $1$ is a constant, so it stays ... um ... constant.)

More generally, we use the algebraic trick your employed by your teacher (and other answers here). I'll present it a little differently. $$\begin{align} \lim_{n\to\infty}\frac{a_pn^p+a_{p-1}n^{p-1} + \cdots + a_0}{b_qn^q+b_{q-1}n^{q-1}+\cdots + b_0} &= \lim_{n\to\infty}\left( \frac{n^p\left(a_p + \dfrac{a_{p-1}}{n}+\cdots+\dfrac{a_0}{n^{p}}\right)}{n^q\left(b_q + \dfrac{b_{q-1}}{n}+\cdots+\dfrac{b_0}{n^q}\right)}\right) \\[6pt] &=\left(\lim_{n\to\infty} \frac{n^p}{n^q}\right)\cdot\left(\lim_{n\to\infty}\frac{a_p + \dfrac{a_{p-1}}{n}+\cdots+\dfrac{a_0}{n^{p}}}{b_q + \dfrac{b_{q-1}}{n}+\cdots+\dfrac{b_0}{n^q}}\right) \\[6pt] &= \left(\lim_{n\to\infty} \frac{n^p}{n^q}\right)\cdot\left(\frac{a_p + 0+\cdots+0}{b_q + 0+\cdots+0}\right) \\[6pt] &= \frac{a_p}{b_q}\;\lim_{n\to\infty}\frac{n^p}{n^q} \\[4pt] &=\begin{cases} \infty & \text{if}\;p > q \\[4pt] a_p/b_q & \text{if}\;p=q \\[4pt] 0 &\text{if}\; p < q \end{cases} \end{align}$$

That is, if the numerator's degree is higher, the limit shoots up to infinity; if the denominator's degree is higher, the limit is dragged down to zero. When the degrees match, the polynomials effectively cancel, leaving only their leading coefficients.

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In the problem at hand, $n^3 + 6 n^2$ has higher degree than $2n^2+5n+6$, so the limit as $n$ approaches infinity should itself be infinity.

Answer 2

Though tempting at first, one should not replace values of the variable inside the limit since this can lead to incorrect results. You can, alternatively, manipulate the expression algebraically:

$$ \lim_{n \to \infty}\frac{n^3+6n^2}{2n^2+5n+6} = \lim_{n \to \infty}\frac{n^3(1+\frac{6}{n})}{n^3(\frac{2}{n}+\frac{5}{n^2}+\frac{6}{n^3})} = \lim_{n \to \infty}\frac{1+\frac{6}{n}}{\frac{2}{n}+\frac{5}{n^2}+\frac{6}{n^3}} $$

I think you can take it from here: the denominator converges to $0$, and the numerator to $1$, thus, the quotient diverges to infinity.

Answer 3

How about this:

$$\lim_{n\to\infty}\dfrac{n^3+6n^2}{2n^2+5n+6} > \lim_{n\to\infty} \dfrac{n^3+6n^2}{6n^2} \Rightarrow \lim_{n\to\infty} \frac{1}{6}n + 1,$$

which diverges to infinity.

Answer 4

Your instructor is correct.

$$\lim\limits_{n\to\infty}\dfrac{n+6}{2+\dfrac{5}{n}+\dfrac{6}{n^2}} =\dfrac{\infty+6}{2+\dfrac{5}{\infty}+\dfrac{6}{\infty^2}}=\dfrac{\infty}{2}=\infty $$

Note that $\infty +6 =\infty $ and $ \frac {\infty}{2}=\infty$