The numerical and graphing techniques discussed so far are great to estimate if a limit exists. To be able to find the true limit value, more algebraic methods are needed. The Limit Laws are the formal mathematical statements that allow limits to be algebraically evaluated intead of using estimation. These laws are outlined on page 70 of the textbook for reference.
Let \(\lim_{x\rightarrow2}\left(f(x)\right)=4,\;\lim_{x\rightarrow2}\left(g(x)\right)=5\) and \(\lim_{x\rightarrow2}\left(h(x)\right)=8\). Use Limit Laws to find \(\lim_{x\rightarrow2}\left(\large{\frac{f(x)-3g(x)}{h(x)}}\right)\).
| Original Problem: | \(\lim_{x\rightarrow2}\left(\large{\frac{f(x)-3g(x)}{h(x)}}\right)\) |
| Quotient Law: | \(\large{\frac{\lim_{x\rightarrow2}\left(f(x)-3g(x)\right)}{\lim_{x\rightarrow2}\left(h(x)\right)}}\) |
| Difference Law: | \(\large{\frac{\lim_{x\rightarrow2}\left(f(x)\right)-\lim_{x\rightarrow2}\left(3g(x)\right)}{\lim_{x\rightarrow2}\left(h(x)\right)}}\) |
| Constant Multiple Law: | \(\large{\frac{\lim_{x\rightarrow2}\left(f(x)\right)-3\lim_{x\rightarrow2}\left(g(x)\right)}{\lim_{x\rightarrow2}\left(h(x)\right)}}\) |
| Plug in limit values: | \(\large{\frac{4-3\cdot5}{8}}\) |
| Simplify: | \(\large{\frac{4-15}{8}}\normalsize{\implies}\frac{-11}{8}\;\color{green}{\checkmark}\) |
These types of functions are defined for all real numbers, except in the rational case, when the denominator is equal to zero. Evaluating limits with these functions can be achieved by directly substituting in the limit value aided by the Limit Laws:
$$\begin{array}{l} \lim_{x\rightarrow a}\left(p(x)\right)=p(a) \\ \\ \lim_{x\rightarrow a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)},\;q(a)\ne0 \end{array}$$Since \(2x^4-8x-16\) is a Polynomial, direct substitution of \(x=4\) will evaluate the limit:
$$\begin{array}{l} 2(4)^4-8(4)-16 \\ \implies 2(256)-32-16 \\ \implies 512-48 \\ \implies 464\;\color{green}{\checkmark} \end{array}$$Since \(\large{\frac{x-1}{x}}\) is a Rational function and \(q(-1)\ne0\), direct substitution of \(x=-1\) will evaluate the limit:
$$\begin{array}{l} \frac{-1-1}{-1} \\ \implies \frac{-2}{-1} \\ \implies 2\;\color{green}{\checkmark} \end{array}$$Consider the following limit: \(\lim_{x\rightarrow5}\left(\large{\frac{2x^2-7x-15}{3x^2-19x+20}}\right)\). The function is a rational function, so it seems direct subsitution would be the method. However, when \(x=5\) is substituted into the denominator: \(3(5)^2-19(5)+20\implies 75-95+20\implies 0\), which makes the entire function undefined at \(x=5\). Since the polynomials are quadratic, a good technique is to try and factor to see if the rational function can be reduced, and then try direct substitution with \(x=5\) again. $$\frac{2x^2-7x-15}{3x^2-19x+20}\implies\frac{(2x+3)(x-5)}{(3x-4)(x-5)}\implies \frac{(2x+3)\color{red}{(x-5)}}{(3x-4)\color{red}{(x-5)}}\implies\frac{2x+3}{3x-4}$$ Now attempting direct substitution again will evaluate the limit:
| Original limit: | \(\lim_{x\rightarrow5}\left(\large{\frac{2x^2-7x-15}{3x^2-19x+20}}\right)\) |
| Simplify: | \(\implies\lim_{x\rightarrow5}\left(\large{\frac{2x+3}{3x-4}}\right)\) |
| Direct substitute \(x=5\): | \(\implies\lim_{x\rightarrow5}\left(\large{\frac{2(5)+3}{3(5)-4}}\right)\) |
| \(\implies\large{\frac{13}{11}}\;\color{green}{\checkmark}\) |
Evaluate the limit: \(\lim_{x\rightarrow9}\left(\large{\frac{\sqrt{x}-3}{x-9}}\right)\).
| Try direct subsitution \(x=9\): | \(\lim_{x\rightarrow9}\left(\large{\frac{\sqrt{9}-3}{9-9}}\right)\normalsize{\implies\frac{0}{0}}\) |
| Multiply by conjugate idea: | \(\large{\frac{\sqrt{x}-3}{x-9}\cdot\color{blue}{\frac{\sqrt{x}+3}{\sqrt{x}+3}}}\normalsize{\implies} \large{\frac{x+3\sqrt{x}-3\sqrt{x}-9}{(x-9)(\sqrt{x}+3)}}\) |
| Simplify and cancel: | \(\large{\frac{\color{blue}{x-9}}{\color{blue}{(x-9)}(\sqrt{x}+3)}}\normalsize{\implies}\large{\frac{1}{\sqrt{x}+3}}\) |
| Now direct substitute \(x=9\): | \(\lim_{x\rightarrow9}\left(\large{\frac{\sqrt{x}-3}{x-9}}\right)\normalsize{\implies\lim_{x\rightarrow9}}\left(\large{\frac{1}{\sqrt{9}+3}}\right)\) |
| \(\large{\frac{1}{3+3}}\normalsize{\implies\frac{1}{6}}\;\color{green}{\checkmark}\) |
NOTE: This conjugate is not a regular algebra technique used. It is the only way to eliminate the zero that comes in the denominator.