The Limit of a Function

The Algebra Method

The algebra method involves algebraically simplifying the function before trying to evaluate its limit.

Often, this simplification just means factoring and dividing, but sometimes more complicated algebraic or even trigonometric steps are needed.

The Table and Graph Methods

To evaluate a limit of a function $\mathrm{f}(\mathrm{x})$ as $\mathrm{x}$ approaches $\mathrm{c}$, the table method involves calculating the values of $\mathrm{f}(\mathrm{x})$ for "enough" values of $\mathrm{x}$ very close to $\mathrm{c}$ so that we can "confidently" determine which value $\mathrm{f}(\mathrm{x})$ is approaching. If $\mathrm{f}(\mathrm{x})$ is well-behaved, we may not need to use very many values for $\mathrm{x}$. However, this method is usually used with complicated functions, and then we need to evaluate $\mathrm{f}(\mathrm{x})$ for lots of values of $\mathrm{x}$.

The graph method is closely related to the table method, but we create a graph of the function instead of a table of values, and then we use the graph to determine which value $f(x)$ is approaching.

Which Method Should You Use?

In general, the algebraic method is preferred because it is precise and does not depend on which values of x we chose or the accuracy of our graph or precision of our calculator. If you can evaluate a limit algebraically, you should do so. Sometimes, however, it will be very difficult to evaluate a limit algebraically, and the table or graph methods offer worthwhile alternatives. Even when you can algebraically evaluate the limit of a function, it is still a good idea to graph the function or evaluate it at a few points just to verify your algebraic answer.

The table and graph methods have the same advantages and disadvantages. Both can be used on very complicated functions which are difficult to handle algebraically or whose algebraic properties you don't know.

Often both methods can be easily programmed on a calculator or computer. However, these two methods are very time-consuming by hand and are prone to round off errors on computers. You need to know how to use these methods when you can't figure out how to use the algebraic method, but you need to use these two methods warily.

Example 4: Evaluate (a) $\lim\limits_{x \rightarrow 0} \frac{x^{2}+5 x+6}{x^{2}+3 x+2}$ and

(b) $\lim\limits_{x \rightarrow-2} \frac{x^{2}+5 x+6}{x^{2}+3 x+2}$

Solution: The function in each limit is the same but $x$ is approaching a different number in each of them.

(a) Since $x \rightarrow 0$, we know that $x$ is getting closer and closer to 0 so the values of the $x^{2}, 5 x$ and $3 x$ terms get as close to 0 as we want. The numerator approaches 6 and the denominator approaches 2, so the values of the whole function get arbitrarily close to $6 / 2=3$, the limit.

(b) As $\mathrm{x}$ approaches $-2$, the numerator and denominator approach 0, and a small number divided by a small number can be almost anything - the ratio depends on the size of the top compared to the bottom. More investigation is needed.

Table Method: If we pick some values of $x$ close to (but not equal to) $-2$, we get the table

Graph Method: The graph of $y=f(x)=\frac{x^{2}+5 x+6}{x^{2}+3 x+2}$ in Fig. 6 shows that the values of $\mathrm{f}(\mathrm{x})$ are very close to $-1$ when the $\mathrm{x}$-values are close to $-2$.

Algebra Method: $f(x)=\frac{x^{2}+5 x+6}{x^{2}+3 x+2}=\frac{(x+2)(x+3)}{(x+2)(x+1)}\text{.}$

We know $x \rightarrow-2$ so $x \neq-2$, and we can divide the top and bottom by $(x+2)$. Then $f(x)=(x+3) /(x+1)$ so $\mathrm{f}(\mathrm{x}) \rightarrow 1 /-1=-1$ as $\mathrm{x} \rightarrow-2$.

$\text { If } \lim\limits_{x \rightarrow c}\left\{\begin{array}{l} \text { polynomial } \\ {\text { polynomial }}\} \text { approaches } \frac{0}{0}, \text { try dividing the top and bottom by } \mathrm{x}-\mathrm{c} \text {. } \end{array}\right.$

Practice 2: (a) $\lim\limits_{x \rightarrow 2} \frac{x^{2}-x-2}{x-2}$   (b) $\lim\limits_{t \rightarrow 0} \frac{t \cdot \sin (t)}{t^{2}+3 t}$   (c) $\lim\limits_{w \rightarrow 2} \frac{w-2}{\ln (w / 2)}$.