Polynomial functions and rational functions.

Proposition 4.5.1   In a neighborhood of $ +\infty$ (resp.$ -\infty $ ), a polynomial function has the same limit as its term with greatest power.

Proof. Let $ P(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+ \dots + a_1x+a_0$ , where all th $ a_i$ 's are real and $ a_n \neq 0$ . Then we have:

$\displaystyle P(x)=a_nx^n \left( 1 + \frac{a_{n-1}}{a_nx}+ \frac{a_{n-2}}{a_nx^...
...s + \frac{a_2}{a_nx^{n-2}}+ \frac{a_1}{a_nx^{n-1}} + \frac{a_0}{a_nx^n} \right)$    

When $ x$ gets either positive or negative with arbitrary large absolute value, every fraction $ \frac{a_k}{a_nx^{n-k}}$ has limit equal to 0, thus the sum in parentheses has limit equal to 1, whence the result. $ \qedsymbol$

Example 4.5.2   For $ f(x)=4x^5-18x^4+7x^4-x+56$ , $ \underset{x \rightarrow +\infty}{\lim}f(x)= \underset{x \rightarrow +\infty}{\lim} 4x^5 = +\infty$ and $ \underset{x \rightarrow -\infty}{\lim}f(x)= \underset{x \rightarrow -\infty}{\lim} 4x^5 = -\infty$ .

The same kind of computation can be useful for non polynomial expressions, as shown in the next example.

Example 4.5.3   Let $ f: \; x \mapsto x^2-5\sqrt{x}-3$ . We compute the limit of $ f$ at $ +\infty$ :

$\displaystyle \forall x \in [0,+\infty), \; f(x)=x^2 \left( 1 -5\frac {\sqrt{x}...
...sqrt{x}}}_{\rightarrow 0} -\underbrace{\frac {3}{x^2}}_{\rightarrow 0} \right).$    

Thus $ \underset{x\rightarrow +\infty}{\lim} f(x)=+\infty$ .

Proposition 4.5.4   In a neighborhood of $ +\infty$ (resp.$ -\infty $ ), a rational function has the same limit as the quotient of the term with greatest power in the numerator by the term with greatest power in the denominator.

The proof is a prototype of the computations that we will make later, for solving other cases where both the numerator and the numerator have an infinite limit at some point $ x_0$ .

Proof. Let $ f$ be a rational function, say

$\displaystyle f(x)=\frac {a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots +a_1x+a_0} {b_mx^m+b_{m-1}x^{m-1}+b_{m-2}x^{m-2}+\dots + b_1x+b_0}.$    

where $ a_n \neq 0$ and $ b_m \neq 0$ . We push out the greatest power of $ x$ in the numerator and the greatest power of $ x$ in the numerator:

$\displaystyle f(x)$ $\displaystyle =\frac{a_nx^n \left( 1+ \frac {a_{n-1}}{a_nx}+ \frac {a_{n-2}}{a_...
...{b_{m-2}}{b_mx^2} + \dots + \frac{b_1}{b_mx^{m-1}}+\frac{b_0}{b_mx^m} \right) }$    
$\displaystyle \quad$ $\displaystyle = \frac {a_nx^{n-m}}{b_m} \; \cdot \; \frac{ 1+ \frac {a_{n-1}}{a...
... \frac {b_{m-2}}{b_mx^2} + \dots + \frac{b_1}{b_mx^{m-1}}+\frac{b_0}{b_mx^m}} .$    

Now every fraction $ \frac {a_{n-r}}{a_nx^r}$ and every fraction $ \frac {b_{m-s}}{b_mx^s}$ has limit equal to 0 when $ x$ gets either positive or negative with arbitrary large absolute value, therefore both expressions in parentheses have a limit at $ \pm \infty$ equal to 1, whence the result. $ \qedsymbol$

Example 4.5.5   If $ f(x)= \frac {3x^3-5x+2}{4x^2-x+6}$ , then:

$\displaystyle \underset{x \rightarrow +\infty}{\lim}f(x)= \underset{x \rightarr...
...c {3x^3}{4x^2} =\underset{x \rightarrow +\infty}{\lim} \; \frac 34 x = +\infty.$    

Noah Dana-Picard 2007-12-28