explanation for the shortcut trick to evaluate $\lim _{x\rightarrow \infty }\dfrac{N}{D}$

Question

This is with reference to this video. If you could not watch the video, I have written the shortcut trick below.

Lets say we have a limit extended to infinity of a polynomial function N by a polynomial function D (or any rational function), where N and D are numerator and denominator respectively; $$\lim _{x\rightarrow \infty }\dfrac{N}{D}$$

Let us consider the degree of the numerator N and the degree of the denominator D. If the degree of the numerator N is greater than the degree of the denominator N then answer is infinity; $$\lim _{x\rightarrow \infty }\dfrac{N}{D}=\infty $$

If in the numerator N the degree is less than the degree of the denominator D then the answer is zero; $$\lim _{x\rightarrow \infty }\dfrac{N}{D}=0 $$

and if the degree of the numerator N and the degree of the denominator D are equal then the answer is coefficient of the highest power in the numerator upon the coefficient of highest power in the denominator. $$\lim _{x\rightarrow \infty }\dfrac{N}{D}=\dfrac{a}{b} $$, where $a$ and $b$ are the coefficient of the highest power in the numerator and the coefficient of highest power in the denominator respectively.

Examples $$ \begin{array}{l} \lim _{x \rightarrow \infty} \frac{x^{4}-3 x^{2}+1}{x^{2}+5}=\infty \\ \lim _{x \rightarrow \infty} \frac{\sqrt{x^{6}-5 x^{3}+2 x+7}}{x^{5}+4 x^{4}-11 x^{3}}=0 \\ \lim _{x \rightarrow \infty} \frac{2 x^{3}-7 x^{4}+2}{5 x^{4}+3 x^{2}+1}=-\frac{7}{5} \end{array} $$

Why does this shortcut trick work? I found this shortcut trick from this video.

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The same shortcut trick has been mentioned by @ParamanandSingh in this answer;

• If the degree of numerator is less than that of the denominator then the limit is $0$.
• If the degree of numerator is equal to that of the denominator then the limit is non-zero and equal to the ratio of leading coefficients of the numerator and denominator.
• If the degree of numerator is more than that of the denominator then the limit is $\infty$ or $-\infty$ depending on whether the ratio of leading coefficients of numerator and denominator is positive or negative.

What is the explanation for this shortcut trick to evaluate $\lim _{x\rightarrow \infty }$? Why does this shortcut trick work?

Answer 1

One of the simpler way to look at this and you can do it as an exercise is to factor out the term with the highest degree for both the numerator and denominator and ponder about it for a second. You will find that essentially (using one of your example) that essentially $$ \lim_{x\to\infty} \frac{x^4-3x^2+1}{x^2+5} = \lim_{x\to\infty} \frac{x^4}{x^2}$$ And this goes for any polynomial with respect to limits as the term with the highest degree grows large.

Answer 2

Let $P(x)=a_mx^m+...+a_0$ and $Q(x)=b_nx^n+...+b_0$ be polynomials with $m,n>0$ and $a_m\ne 0\ne b_n.$

For $x\ne 0$ we have $P(x)=a_mx^mP_1(x)$ and $Q(x)=b_nx^nQ_1(x)$ where

$P_1(x)=1+a_m^{-1}a_{m-1}x^{-1}+...+a_m^{-1}a_0x^{-m}$

and

$Q_1(x)=1+b_n^{-1}b_{n-1}x^{-1}+...+b_n^{-1}b_0x^{-n}.$

Observe that $P_1(x)$ and $Q_1(x)$ both converge to $1$ as $|x|\to \infty.$ And observe that for $x\ne 0$ we have $$\frac {P(x)}{Q(x)}=\frac {a_mx^m}{b_nx^n} \cdot\frac {P_1(x)}{Q_1(x)}.$$