Limits of polynomials and rational functions

Part 3 · Limits of functions and continuity · Chapter 3 · lecture notes by Fabio Furini · Chapter PDF

1. Limits of polynomials at \(\pm \infty\)

• A polynomial of (maximum) degree \(n\) can be written as:

  \[ P_n(x) = \sum_{i=0}^n a_i \: x^i \quad {\rm ~~with~~} a_i \in \R, {\rm ~for~~} i \in \{0,1,\dots,n\}, ~ {\rm ~~and~~} a_n \neq 0. \]

  The value \(a_i\) is the coefficient of the monomial \(i\), with \(i=0,1,\dots,n\), while \(x^i\) is the literal part of the monomial. In this notation, without loss of generality, the monomials are ordered by increasing values of the exponents. ¹

• Factoring out \(a_n \: x^n\), that is, the last monomial, which is the one of maximum degree, we have:

  \[ P_n(x) = a_n \: x^n \left( \overbrace{\frac{a_0}{a_n \: x^n} + \frac{a_1}{a_n \: x^{n-1}} + {\rm \dots} + \frac{a_{n-1}}{a_n \: x}}^{\rr 0 {\rm ~~as~~} x \rr \pm \infty} + 1 \right) \]

\[ \lim_{x \rr \pm \infty} P_n(x) = \lim_{x \rr \pm \infty} a_n \: x^n \]

To compute the limit of a polynomial as \(x \rr \pm \infty\), it is enough to compute the limit of the monomial of maximum degree.

Example 1: Limits of polynomials as \(x \rr \pm \infty\)

For example:

\[ \lim_{x \rr \ip} (8 - x^2 + 3\:x^3)= \lim_{x \rr \ip} 3\:x^3 \: \left( \underbrace{\frac{8}{3\:x^3} - \frac{1}{3\:x}}_{\rr 0 {\rm ~as~} x \rr \ip } + 1 \right) = \lim_{x \rr \ip} 3\:x^3 = \ip \]

\[ \lim_{x \rr \im} (120 + 4\:x -2\:x^3) = \lim_{x \rr \im} -2\:x^3 \: \left( \underbrace{\frac{120}{-2\:x^3} + \frac{4}{-2\:x^2}}_{\rr 0 {\rm ~as~} x \rr \im} + 1 \right) = \lim_{x \rr \im} -2\:x^3 = \ip \]

2. Limits of rational functions at \(\pm \infty\)

• Let \(f\) be a rational function (ratio of polynomials):

  \[ f(x) = \frac{P_n(x)}{P_m(x)} \]

  where \(P_n(x)\) and \(P_m(x)\) are polynomials of degree \(n\) and \(m\), respectively:

  \[ P_n(x) = \sum_{i=0}^n a_i \: x^i \quad {\rm ~~with~~} a_i \in \R, {\rm ~for~~} i=0,1,\dots,n, ~ {\rm ~~and~~} a_n \neq 0. \]
  \[ P_m(x) = \sum_{i=0}^m b_i \: x^i \quad {\rm ~~with~~} b_i \in \R, {\rm ~for~~} i=0,1,\dots,m, ~ {\rm ~~and~~} b_m \neq 0. \]

• Factoring out \(a_n \: x^n\) in the numerator and \(b_m \: x^m\) in the denominator, that is, the monomials of maximum degree, we have:

  \[ \frac{P_n(x)}{P_m(x)} = \frac{a_n \: x^n}{b_m \: x^m} \frac{\left \{ \overbrace{\frac{a_0}{a_n \: x^n} + \frac{a_1}{a_n \: x^{n-1}} + \dots + \frac{a_{n-1}}{a_n \: x}}^{\rr 0 {\rm ~~as~~} x \rr \pm \infty} + 1 \right\}}{\left \{ \underbrace{\frac{b_0}{b_m \: x^m} + \frac{b_1}{b_m \: x^{m-1}} + \dots + \frac{b_{m-1}}{b_m \: x}}_{\rr 0 {\rm ~~as~~} x \rr \pm \infty} + 1 \right\}}. \]

\[ \lim_{x \rr \pm \infty} \frac{P_n(x)}{P_m(x)} = \lim_{x \rr \pm \infty} \frac{a_n \: x^n}{b_m \: x^m} = \lim_{x \rr \pm \infty} \frac{a_n }{b_m} \: x^{n-m} \]

To compute the limit of a ratio of polynomials as \(x \rr \pm \infty\), it is enough to compute the limit of the ratio of the monomials of maximum degree.

Example 2: Limits of rational functions as \(x \rr \pm \infty\)

For example:

\[ \lim_{x \rr \im} \frac{1+3\: x^3}{2 + x^2} = \lim_{x \rr \im} \frac{3\: x^3}{x^2} =\lim_{x \rr \im} 3\:{x} = \im \]

\[ \lim_{x \rr \ip} \frac{1+3\: x^3}{1- 10x +x^4 } = \lim_{x \rr \ip} \frac{3\: x^3}{x^4} = \lim_{x \rr \ip} 3\: \frac{1}{x} = 0 \]

\[ \lim_{x \rr \im} \frac{ 7\: x^2 + 6\: x^5}{-4+7 \: x^5} =\lim_{x \rr \im} \frac{6\: x^5}{7 \: x^5}= \lim_{x \rr \im} \frac{6}{7} = \frac{6}{7} \]

To compute the limit of rational functions (without the constant term) as \(x \to 0\) we must factor out, in the numerator and in the denominator, the powers of minimum degree.

Example 3: Limits of rational functions as \(x \rr 0\)

For example:

\[ \lim_{x \rr 0} \frac{-2x^2+3\: x^3-6\: x^5}{x+x^2 - x^7} = \lim_{x \rr 0} \frac{ -2x^2 \left(1 \overbrace{-\frac{3}{2}\:x+ 3 \:x^3}^{\to 0 {\rm~as~} x \to 0} \right)}{ x \left(1 \underbrace{+x- \:x^6}_{\to 0 {\rm~as~} x \to 0} \right)} = \lim_{x \rr 0} -2\:x = 0 \]

3. Limits of quotients of sums of powers with rational exponents

• The same rules hold for quotients of sums of powers with rational exponents:

  1. as \(x \rr \infty\) we must factor out the power with the maximum exponent

  2. as \(x \rr 0^+\) we must factor out the power with the minimum exponent

Raising to a power can also be defined with a negative base if the exponent is a rational number (fraction) with an odd denominator. Hence in this case we can compute the limit as \(x \rr 0\) (otherwise only as \(x \rr 0^+\)).

Example 4: Limits of quotients of sums of powers with rational exponents

For example:

\[ \lim_{x \rr 0} \frac{x^{2/3}}{ 3\: {x}^{1/3} + x + x^2 } \]

Factoring out in the denominator the power with the minimum exponent, we have:

\[ \frac{x^{2/3}}{ 3\: x^{1/3} + x + x^2} = \frac{x^{2/3}}{3 \: x^{1/3} \left(1+ \frac{x}{3\:x^{1/3}} + \frac{x^2}{3\:x^{1/3}} \right)} = \frac{x^{2/3}}{3 \: x^{1/3} \left(1+ \frac{x^{2/3}}{3} + \frac{x^{5/3}}{3} \right)} = \frac{1}{3} \: x^{1/3} \frac{1}{\left(1+ \frac{x^{2/3}}{3} + \frac{x^{5/3}}{3} \right)} \]

hence

\[ \lim_{x \rr 0} \frac{x^{2/3}}{ 3\: {x}^{1/3} + x + x^2} = \frac{1}{3} \: \lim_{x \rr 0} \: x^{1/3} \frac{1}{\left(1+ \underbrace{\frac{x^{2/3}}{3} + \frac{x^{5/3}}{3}}_{\rr 0 {\rm ~~as~~} x \rr 0} \right)} = 0 \]

Another option to compute limits as \(x \rr 0\) is to make a change of variable:

\[ x=\frac{1}{y}, {\rm ~~if~~} x \rr 0^{+} {\rm ~~then~~} y \rr \ip, {\rm ~~if~~} x \rr 0^{-} {\rm ~~then~~} y \rr \im \]

and hence we reduce to the case of limits at \(\pm\infty\).

Example 5: Computing the limit of rational functions as \(x \rr 0^+\) with a change of variable

We want to compute the following limit:

\[ \lim_{x \rr 0} \frac{ x^{2/3} }{ 3\: x^{1/3} + x + x^2 }, \qquad f(x)=\frac{ x^{2/3} }{ 3\: x^{1/3} + x + x^2 } \]

Making the change of variable \(x=\frac{1}{y}\), we therefore have to compute the following two limits:

\[ \lim_{x \rr 0^+} f(x) = \lim_{y \rr \ip} \frac{ \left( \frac{1}{y} \right)^{2/3} }{ 3\: \left( \frac{1}{y} \right)^{1/3} + \frac{1}{y} + \left( \frac{1}{y} \right)^2 } {\rm ~~~~and~~~~} \lim_{x \rr 0^-} f(x) = \lim_{y \rr \im} \frac{ \left( \frac{1}{y} \right)^{2/3} }{ 3\: \left( \frac{1}{y} \right)^{1/3} + \frac{1}{y} + \left( \frac{1}{y} \right)^2 } \]

We have:

\[ \frac{ \left( \frac{1}{y} \right)^{2/3} }{ 3\: \left( \frac{1}{y} \right)^{1/3} + \frac{1}{y} + \left( \frac{1}{y} \right)^2 } = \frac{y^{-2/3}}{3\:y^{-1/3} + y^{-1} + y^{-2}} \]

Factoring out in the denominator the power with the maximum exponent, we have:

\[ \frac{y^{-2/3}}{3\:y^{-1/3} + y^{-1} + y^{-2}} = \frac{y^{-2/3}}{3\:y^{-1/3} \left(1 + \frac{y^{-1}}{3\:y^{-1/3}}+ \frac{y^{-2}}{3\:y^{-1/3}} \right)} = \frac{1}{3} \: \frac{1}{ y^{1/3}} \frac{1}{\left(1 + \frac{1}{3\:y^{2/3}}+ \frac{1}{3\:y^{5/3}} \right)} \]

Hence:

\[ \frac{1}{3} \: \lim_{y \rr \im} \frac{1}{ y^{1/3}} \frac{1}{\left(1 + \frac{1}{3\:y^{2/3}}+ \frac{1}{3\:y^{5/3}} \right)} = \frac{1}{3} \: \lim_{y \rr \ip} \frac{1}{ y^{1/3}} \frac{1}{\left(1 + \frac{1}{3\:y^{2/3}}+ \frac{1}{3\:y^{5/3}} \right)} = 0 \]

since, as \(y \rr \pm \infty\), we have

\[ \frac{1}{y^{1/3}} \rr 0 {\rm ~~and~~} \left(1 + \frac{1}{3\:y^{2/3}}+ \frac{1}{3\:y^{5/3}} \right) \rr 1 {\rm ~~~~since~~~} \frac{1}{3\:y^{2/3}} \rr 0, ~~\frac{1}{3\:y^{5/3}} \rr 0 . \]

Since the right limit and the left limit of \(f(x)\) as \(x \rr 0\) exist and are equal to \(0\), then:

\[ \lim_{x \rr 0} f(x) =0 \]

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1. If a monomial of degree \(i\) is missing we have \(a_i=0\); \(a_0\) is the constant term since \(x^0=1.\) ↩