09Bmorelimits

= Limits of Discontinuous Functions =


 * If a function is discontinuous, such as the one shown here:


 * then the limit becomes interesting at the point where the graph is discontinous


 * For example, in the hybrid function shown:
 * the limit as x approaches 0 from the negative side is 2
 * left limit = 2
 * the limit as x approaches 0 from the positive side is 0
 * right limit = 0


 * In notation:

math . \qquad g \big( x \big) = \left\{ \begin{matrix} x + 2 & x \leqslant 0 \\ \\ x^2 & x > 0 \\ \end{matrix} \right.\\.\\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 0^- \\ \end{matrix} \;\; g(x) = 2 \\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 0^+ \\ \end{matrix} \;\; g(x) = 0 math


 * The notation x → a – means x approaches a from the negative side (left limit)
 * not to be confused with x → – a (x approaches negative a)


 * Clearly, at x = 0, the left limit of g(x) is not equal to the right limit of g(x)


 * When a Limit Exists (or is Defined) **


 * ** A limit at a point only exists **** when **(is only defined when)
 * ** the left limit = the right limit **
 * the function does __**not**__ have to be defined at that point


 * Example 1 **


 * in the function ( h(x) ) shown here
 * at the point x = 2
 * the left limit = 1
 * the right limit = 1


 * so at x = 2, the limit exists and is equal to 1


 * In notation:

math . \qquad h \big( x \big) = \left\{ \begin{matrix} \big(x-3\big)^2 & x \leqslant 2 \\ \\ x-1 & x > 2 \\ \end{matrix} \right.\\.\\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 2^- \\ \end{matrix} \;\; h(x) = 1 \\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 2^+ \\ \end{matrix} \;\; h(x) = 1 math

math . \qquad \qquad \text{so} \\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 2^- \\ \end{matrix} h \big( x \big) = \begin{matrix} \lim \\ x \to 2^+ \\ \end{matrix} h \big( x \big) \\.\\.\\ . \qquad \qquad \text{hence} \\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 2 \\ \end{matrix} h(x) = 1 math


 * Example 2 **
 * in the function shown here

math . \qquad \qquad y = x + 1 \qquad x \in R \; \backslash \big\{ 2 \big\} math


 * at the point x = 2
 * left limit = 3
 * right limit = 3


 * so at x = 2 the limit exists and is equal to 3
 * even though the function is not defined at x = 2

Limits of Rational Functions


 * A ** Rational Function ** consists of a fraction where both the numerator and denominator are polynomials
 * remember that a normal number counts as a polynomial of degree 0


 * Examples of rational functions include

math . \qquad \qquad y = \dfrac{2}{x+2} \\.\\ . \qquad \qquad f(x) = \dfrac{3x-1}{x^2+4} \\.\\ . \qquad \qquad g(x) = \dfrac{x - 5x}{4} math


 * The maximal domain of any rational function is the set of real numbers (R) excluding values where the denominator equals zero


 * If asked to find the limit involving a rational function, substitute the limiting value into the function
 * if you can get a result, the result is the limit


 * If you can't get a result because the limiting value makes the denominator equal to zero
 * you may still be able to find the limit by factorising and cancelling the rational function
 * then substituting the limiting value into the simplified form of the function


 * Example 3 **

math . \qquad \text{For the function} : \quad f \big( x \big) = \dfrac{x^2-x-2}{x-2} math

math . \qquad \text{Find } \\.\\ . \qquad \big( \textbf{a} \big) \quad \begin{matrix} \lim \\ x \to 4 \\ \end{matrix} \;\; f(x) \\.\\ . \qquad \big( \textbf{b} \big) \quad \begin{matrix} \lim \\ x \to 2 \\ \end{matrix} \;\; f(x) math


 * Solution: **

... ... ** (a) **

math . \qquad \qquad \begin{matrix} \lim \\ x \to 4 \\ \end{matrix} \;\; \left( \dfrac{x^2-x-2}{x-2} \right) \\.\\ . \qquad \qquad \qquad = \dfrac{ 4^2-4-2}{4-2} \\.\\ . \qquad \qquad \qquad = \dfrac{10}{2} \\.\\ . \qquad \qquad \qquad = 5 math

math . \qquad \text{so} \\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 4 \\ \end{matrix} \;\; f(x) = 5 math

... ... ** (b) **

math . \qquad \qquad \begin{matrix} \lim \\ x \to 2 \\ \end{matrix} \;\; \left( \dfrac{x^2-x-2}{x-2} \right) \\.\\ . \qquad \qquad \qquad = \dfrac{ 2^2-2-2}{2-2} \\.\\ . \qquad \qquad \qquad = \dfrac{0}{0} \\.\\ . \qquad \qquad \qquad = \textit{undefined} math
 * ** BUT **, we try factorising and simplifying the fraction, then substitute again

math . \qquad \qquad \begin{matrix} \lim \\ x \to 2 \\ \end{matrix} \;\; \left( \dfrac{x^2-x-2}{x-2} \right) \\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 2 \\ \end{matrix} \;\; \left( \dfrac{ (x-2)(x+1) }{x-2} \right) \\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 2 \\ \end{matrix} \;\; \Big( x+1 \Big) \\.\\ . \qquad \qquad \qquad = 2 + 1 \\.\\ . \qquad \qquad \qquad = 3 math

math . \qquad \text{so} \\.\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 2 \\ \end{matrix} \;\; f(x) = 3 math


 * the limit at x = 2 exists, even though f(2) is undefined.
 * the graph of f(x) is shown at the right.
 * it is equivalent to f(x) = x + 1 .. for x Î R \ {2}


 * Note **
 * Not all rational functions can be simplified in this way to allow us to find a limit
 * For some rational functions, the limit for a particular value may be undefined.
 * For example, consider the hyperbola

math . \qquad \qquad g(x) = \dfrac{1}{x - 3} \\.\\ .\\ . \qquad \qquad \begin{matrix} \lim \\ x \to 3 \\ \end{matrix} \;\; g(x) = \textit{ undefined} math


 * this is because x = 3 is a vertical asymptote on the hyperbola


 * at x = 3
 * left limit = – ¥
 * right limit = + ¥

.
 * so the limit on g(x) at x = 3 does not exist