Algebraically Finding a Limit « The Science That Draws Necessary Conclusions

Algebraically Finding a Limit

Finding a limit may be as simple as plugging a number into a function, or it may require manipulating the function if it is undefined at that point. $f(x) = 3x +1$ is a linear function continuous at all points. The limit as x approaches any point a is f(a), $\displaystyle \lim_{x \to a}f(x) = f(a)$ . To find the limit as x approaches 2 in f(x), just plug x into f(x), $\displaystyle\lim_{x \to 2}f(x)=f(2)=3(2)+1=7$ .

Now look at the rational function $g(x)=\frac{x+1}{x-1}$ . g(x) is continuous at every point except when x=1. In order to solve the limit as x approaches 1, g(x) must be manipulated to defined at 1. Multiplying g(x) by $\frac{x-1}{x-1}$ will return the exact same function with the exception of g(1) being defined, (x+1)(x-1). The limit can be taken using this function so $\displaystyle \lim_{x \to 1}g(x)=\lim_{x \to 1}(x+1)(x-1)=0$ .

Another good example is $\displaystyle\lim_{x \to \pm 2}h(x)$ if $h(x)=\frac{x+2}{x^2-4}$ . h(x) is undefined at -2 and 2 so the function needs to be manipulated to be defined at those two points. $x^2-4$ is the difference of two squares so it can be factored to $(x-2)(x+2)$ . The numerator and denominator both share the factor (x+2) so it can be cancelled leaving the function $\frac{1}{x-2}$ . Now it is clear that $\displaystyle\lim_{x \to -2}h(x)=\lim_{x \to -2}\frac{1}{x-2}=-\frac{1}{4}$ ; however the new function is still undefined at 2. Because the function cannot be simplified any more, the limit as x approaches 2 does not exits, $\displaystyle\lim_{x \to 2}h(x) = DNE$ .