Limits of Rational Functions

Men lie, women lie, numbers don’t, Lil B

Recall

Definition. A function f is a rule, relationship, or correspondence that assigns to each element of one set (x ∈ D), called the domain, exactly one element of a second set, called the range (y ∈ E).

The pair (x, y) is denoted as y = f(x). Typically, the sets D and E will be both the set of real numbers, ℝ. A mathematical function is like a black box that takes certain input values and generates corresponding output values (Figure E).

Very loosing speaking, a limit is the value to which a function grows close as the input get closer and closer to some other given value.

One would say that the limit of f, as x approaches a, is L, $\lim_{x \to a} f(x)=L$. Formally, for every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ implies that | f(x) − L | < ε. In other words, f(x) gets closer and closer to L, f(x)∈ (L-ε, L+ε), as x moves closer and closer -approaching closer but never touching- to a (x ∈ (a-δ, a+δ), x≠a)) -Fig 1.a.-

Definition. Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a, then we say that, $\lim_{x \to a} f(x) = L$ if

$\forall \epsilon>0, \exists \delta>0: 0<|x-a|<\delta, implies~ |f(x)-L|<\epsilon$

Or

$\forall \epsilon>0, \exists \delta>0: |f(x)-L|<\epsilon, whenever~ 0<|x-a|<\delta$   

Definition of Rational Functions

Definition. Rational functions are ratios of two polynomial functions, f(x) = $\frac{p(x)}{q(x)} = \frac{a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0}{b_mx^m+b_{m−1}x^{m−1}+…+b_1x+b_0}$ where a_n ≠ 0, b_m ≠ 0, and q(x) ≠ 0, e.g., $\frac{3-2x}{x-2}, \frac{x^3 + x^2 - 2x + 12}{x+3}.$

Limits of Rational Functions.

Let’s try to calculate $\lim_{x \to a} f(x)$:

• Direct substitution. If a ∈ Dom(f), we substitute or plug the value into the expression.

1. $\lim_{x \to 1} \frac{3-2x}{x-2} = \frac{3-2}{1-2} = -1$.
2. $\lim_{x \to 2} \frac{4x}{2x-3} = \frac{4·2}{4-3} = 8.$

• If a ∉ Dom(f) and substitution produces a zero in both denominator and numerator -indeterminate form-, it sometimes works to factor and then cancel.

1. $\lim_{x \to -4} \frac{x^2+3x-4}{x+4} = \lim_{x \to -4} \frac{(x-1)(x+4)}{x+4} = \lim_{x \to -4} (x-1) = -5$
2. $\lim_{x \to 3} \frac{x^2-x-6}{x-3} = \lim_{x \to 3} \frac{(x+2)(x-3)}{x-3} = \lim_{x \to 3} (x+2) = 5$
3. $\lim_{x \to -3} \frac{x^3 + x^2 - 2x + 12}{x+3} = \lim_{x \to -3} \frac{(x+3)(x^2-2x+4)}{x+3} = \lim_{x \to -3} (x^2-2x+4) = 9+6+4=19$

• When the expression inside the limit contains a sum or difference of fractions, it sometimes works to combine the fractions.

1. $\lim_{x \to 2} \frac{\frac{1}{x}-\frac{1}{2}}{x-2} = \lim_{x \to 2} \frac{\frac{2-x}{2x}}{x-2} = \lim_{x \to 2} \frac{-1}{2x} = \frac{-1}{4}$
2. $\lim_{x \to 2}(\frac{1}{4x-8}-\frac{1}{x^2-4}) = \lim_{x \to 2} (\frac{1}{4(x-2)}-\frac{1}{(x-2)(x+2)}) = \lim_{x \to 2} (\frac{(x+2) - 4}{4(x-2)(x+2)}) = \lim_{x \to 2} \frac{x-2}{4(x-2)(x+2)} = \lim_{x \to 2} \frac{1}{4(x+2)} = \frac{1}{16}.$
3. $\lim_{x \to 2} \frac{4}{x^2-4}-\frac{1}{x-2}$=[∞-∞] $\lim_{x \to 2} \frac{4-(x+2)}{(x-2)(x+2)} = \lim_{x \to 2} \frac{(2-x)}{(x-2)(x+2)} = \lim_{x \to 2} \frac{-1}{(x+2)} = \frac{-1}{4}.$

• The conjugate method (Rationalization). When the expression inside the limit contains a radical expression, it sometimes works to rationalize, that is, multiply numerator and denominator by a radical that will get rid of the radical in the denominator (or numerator).

1. $\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x} = \lim_{x \to 0} \frac{\sqrt{1+x}-1}{x}·\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} = \lim_{x \to 0} \frac{(1+x)-1}{x(\sqrt{1+x}+1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{1+x}+1)} = \lim_{x \to 0} \frac{1}{(\sqrt{1+x}+1)} = \frac{1}{2}$.
2. $\lim_{x \to 0} (\frac{3}{x\sqrt{9-x}}-\frac{1}{x})$[Combine the fractions] $\lim_{x \to 0} (\frac{3-\sqrt{9-x}}{x\sqrt{9-x}}) = \lim_{x \to 0} (\frac{3-\sqrt{9-x}}{x\sqrt{9-x}}·\frac{3+\sqrt{9-x}}{3+\sqrt{9-x}}) = \lim_{x \to 0} \frac{9-(9-x)}{(x\sqrt{9-x})(3+\sqrt{9-x})} = \lim_{x \to 0} \frac{x}{(x\sqrt{9-x})(3+\sqrt{9-x})} = \lim_{x \to 0} \frac{1}{(\sqrt{9-x})(3+\sqrt{9-x})} = \frac{1}{\sqrt{9}(3+\sqrt{9})} = \frac{1}{3·6} = \frac{1}{18}$
3. $\lim_{x \to 2} \frac{3-\sqrt{2x+5}}{x-2} = \lim_{x \to 2} \frac{3-\sqrt{2x+5}}{x-2}\frac{3+\sqrt{2x+5}}{3+\sqrt{2x+5}} = \lim_{x \to 2} \frac{9-(2x+5)}{(x-2)(3+\sqrt{2x+5})} = \lim_{x \to 2} \frac{-2x+4}{(x-2)(3+\sqrt{2x+5})} = \lim_{x \to 2} \frac{-2(x-2)}{(x-2)(3+\sqrt{2x+5})} = \lim_{x \to 2} \frac{-2}{(3+\sqrt{2x+5})} = \frac{-2}{3+\sqrt{9}} = \frac{-2}{6} = \frac{-1}{3}$
4. $\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3} = \lim_{x \to 9} \frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)} = \lim_{x \to 9} \frac{(x-9)(\sqrt{x}+3)}{x-9} = \lim_{x \to 9} \sqrt{x}+3 = \sqrt{9} + 3 = 6.$
5. $\lim_{x \to 0} \frac{1}{x\sqrt{x+1}}-\frac{1}{x} =$ [∞-∞] $\lim_{x \to 0} \frac{1-\sqrt{x+1}}{x\sqrt{x+1}} = \lim_{x \to 0} \frac{1-\sqrt{x+1}}{x\sqrt{x+1}} \frac{1+\sqrt{x+1}}{1+\sqrt{x+1}} = \lim_{x \to 0} \frac{1-(x+1)}{x\sqrt{x+1}(1+\sqrt{x+1})} = \lim_{x \to 0} \frac{-x}{x\sqrt{x+1}(1+\sqrt{x+1})} = \lim_{x \to 0} \frac{-1}{\sqrt{x+1}(1+\sqrt{x+1})} = \frac{-1}{2}$

Limits at infinite

The value of $\lim_{x \to ∞} f(x)$ can be determined by dividing the numerator and denominator by the highest power of x appearing in the denominator. This determines which term(s) in the overall expression dominate(s) the behavior of the function at large values of x.

$\lim_{x \to ∞} \frac{2x^2}{(x^2+1)(x-3)} = \lim_{x \to ∞}\frac{2x^2}{x^3-3x^2+x-3}$ =[Apply L’Hôpital’s rule or divide by x³] = $\lim_{x \to ∞}\frac{\frac{2}{x}}{1-3\frac{1}{x}+\frac{1}{x^2}-3\frac{1}{x^3}} = 0$.

$\lim_{x \to -∞} \frac{7x^3-x+2}{2x^2-5x-6}$ =[Apply L’Hôpital’s rule or divide by x²] = $\lim_{x \to -∞}\frac{7x-\frac{1}{x}+\frac{2}{x^2}}{2-\frac{5}{x}-\frac{6}{x^2}} = -∞$.

$\lim_{x \to -∞} \frac{7x^3-x+2}{2x^3-5x-6}$ =[Apply L’Hôpital’s rule or divide by x³] = $\lim_{x \to -∞}\frac{7-\frac{1}{x^2}+\frac{2}{x^3}}{2-\frac{5}{x^2}-\frac{6}{x^3}} = \frac{7}{2}$.

The limits at infinity for a rational function, say f(x) = $\frac{p(x)}{q(x)} = \frac{a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0}{b_mx^m+b_{m−1}x^{m−1}+…+b_1x+b_0}$ can be exclusively determined or calculated based on its degrees:

• The degree of the numerator and the denominator are the same, $\lim_{x \to ∞} f(x) = \lim_{x \to ∞} \frac{a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0}{b_mx^m+b_{m−1}x^{m−1}+…+b_1x+b_0} = \frac{a_n}{b_m}$ where m = n, a_n and b_n = b_m are the leading coefficients of p and q respectively. The line $ y = \frac{a_n}{b_m}$ is a horizontal asymptote.

1. $\lim_{x \to ∞} \frac{2x^7+4x^3+2x+1}{3x^7+4x^2+3x+5} = \lim_{x \to ∞} \frac{2x^7}{3x^7} = \lim_{x \to ∞} \frac{2}{3} = \frac{2}{3}.$
2. $\lim_{x \to ∞} \frac{4x^2+2x+7}{3x^2+3x+2} = \frac{4}{3}.$
3. $\lim_{x \to ∞} \frac{2x^5+3x^4+2x^3+7x+1}{2x^5+12x^4+3x^2+2x+8} = 1.$

• The degree of the numerator is smaller than the degree of the denominator, $\lim_{x \to ∞} f(x) = \lim_{x \to ∞} \frac{a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0}{b_mx^m+b_{m−1}x^{m−1}+…+b_1x+b_0} = 0$. The line y = 0 is a horizontal asymptote.

1. $\lim_{x \to ±∞} \frac{2x^6+4x^3+2x+1}{3x^7+4x^2+3x+5} = \lim_{x \to ±∞} \frac{2x^6}{3x^7} = \lim_{x \to ±∞} \frac{2}{3x} = 0.$
2. $\lim_{x \to ∞} \frac{4x^2+2x+7}{3x^3+3x+2} =0.$
3. $\lim_{x \to ∞} \frac{2x^5+3x^4+2x^3+7x+1}{2x^7+2x^4+3x^2+2x+8} = 0.$

• The degree of the numerator is bigger than the degree of the denominator, $\lim_{x \to ∞} f(x) = \lim_{x \to ∞} \frac{a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0}{b_mx^m+b_{m−1}x^{m−1}+…+b_1x+b_0} = ±∞$.

1. $\lim_{x \to ∞} \frac{2x^5+8x^2+8}{9x^3+4x^2+3x+5} = \lim_{x \to ∞} \frac{2x^5}{9x^3} = \lim_{x \to ∞} \frac{2x^2}{9} = ∞.$
2. $\lim_{x \to -∞} \frac{4x^3+2x+7}{3x^2+3x+2} = \lim_{x \to -∞} \frac{4x^3}{3x^2} = \lim_{x \to -∞} \frac{4x}{3} = -∞.$
3. $\lim_{x \to ∞} \frac{4x^3+2x+7}{3x^2+3x+2} = \lim_{x \to ∞} \frac{4x^3}{3x^2} = \lim_{x \to ∞} \frac{4x}{3} = ∞.$

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