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 x in a set D, x ∈ D (called the domain) exactly one element y in a set E, y ∈ E (called the codomain or range). 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 mathematical function gets closer and closer to as the input gets closer and closer to some given value.

A limit describe what is happening around a given point, say “a”. It is the value that the function approaches as the input approaches “a”, and it does not depend on the actual value of the function at a, or even on whether the function is defined at “a” at all.

Limits are essential to calculus and mathematical analysis and the understanding of how functions behave. The concept of a limit can be written or expressed as $\lim_{x \to a} f(x) = L.$ This notation is read as “the limit of f as x approaches a equals L”.

Intuitively, this means that the values of f(x) can be made arbitrarily close to L (and I mean as close as we like, e.g., L ± 0.1, L ± 0.01, L ± 0.001, and so on), by choosing values of x sufficiently close to a, but not necessarily equal to a.

Formal definition. We say that the limit of f, as x approaches a, is L, and write $\lim_{x \to a}f(x) = L$. For every real ε > 0, there exists a real δ > 0 such that whenever 0 < | x − a | < δ we have | f(x) − L | < ε. In other words, we can make f(x) arbitrarily close to L, f(x)∈ (L-ε, L+ε) (within any distance ε > 0) by making x sufficiently close to a (within some distance δ > 0, but not equal to a) (x ∈ (a-δ, a+δ), x ≠ a) -Fig 1.a.-

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. $\lim_{x \to 0} \frac{1}{x\sqrt{x+1}}-\frac{1}{x} =$ [∞-∞] $\lim_{x \to 0} \frac{1-\sqrt{x+1}}{x\sqrt{x+1}} = \lim_{x \to 0} \frac{1-\sqrt{x+1}}{x\sqrt{x+1}} \frac{1+\sqrt{x+1}}{1+\sqrt{x+1}} = \lim_{x \to 0} \frac{1-(x+1)}{x\sqrt{x+1}(1+\sqrt{x+1})} = \lim_{x \to 0} \frac{-x}{x\sqrt{x+1}(1+\sqrt{x+1})} = \lim_{x \to 0} \frac{-1}{\sqrt{x+1}(1+\sqrt{x+1})} = \frac{-1}{2}$
6. $\lim_{x \to 3}\frac{\sqrt{x+1}-2}{x-3} = \lim_{x \to 3}\frac{\sqrt{x+1}-2}{x-3}\cdot \frac{\sqrt{x+1}+2}{\sqrt{x+1}+2} = lim_{x \to 3} \frac{x + 1 -4}{(x -3)(\sqrt{x+1}+2)} = lim_{x \to 3} \frac{x -3}{(x -3)(\sqrt{x+1}+2)} = lim_{x \to 3} \frac{1}{\sqrt{x+1}+2} = \frac{1}{4}$

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

Bibliography

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