Given the function \[g(t) = \dfrac{4(t^2- 4)}{t - 2}\] what is the value of $g(t)$ when $t$ approaches $2$?
The function $f$ has the limit $L$ as $x$ approaches $a$, written \[\lim_{x\rightarrow a}f(x) = L\] if the value of $f(x)$ can be made as close to the number $L$ as we please by taking $x$ sufficiently close to but not equal to $a$.
Important: the limit is not equivalent to plugging in / evaluation at $x = a$.
We saw evaluating values closer and closer to the value we are interested in is one technique in finding limits. We can also use the graph.
Let $f(x) = x^3$. Find $\displaystyle\lim_{x \rightarrow 2}f(x)$.
Let \[g(x) = \begin{cases} x + 2 & x \neq 1 \\ 1 & x = 1 \end{cases}\]
Find $\lim_{x\rightarrow 1} g(x)$.
Find the limit of the following functions at the given value of $x$:
$f(x) = \begin{cases} -1 & x < 0 \\ 1 & x \geq 0\end{cases} \ $ at $x = 0$.
$\displaystyle g(x) = \frac{1}{x^2}$ at $x = 0$.
Be careful of using the table and graph. Why:
Find the limit of \[f(x) = \cos(x) + \frac{1}{10000}\] at $x = 0$.
We know how to use a table and the graph to find a limit. Let's see some algebraic properties of limits.
Properties of Limits
Suppose \[\lim_{x\rightarrow a} f(x) = L \qquad \lim_{x\rightarrow a} g(x) = M\] and $r$ is a positive constant, $c \in \mathbb{R}$. Then
From the graphs we can see \[\lim_{x \rightarrow -2} f(x) = 1 \qquad \lim_{x \rightarrow -2} g(x) = -1\]
So by using the limit laws we see \begin{align}\lim_{x\rightarrow -2} [f(x) + 5g(x)] &= \lim_{x\rightarrow -2} f(x) + \lim_{x\rightarrow -2} [5g(x)]
\\&= \lim_{x\rightarrow -2} f(x) + 5\lim_{x \rightarrow -2}g(x)
\\&= 1 + 5\cdot (-1) = -4\end{align}
Here are five more limit laws that are extremely useful.
More Properties of Limits
Suppose $a, c \in \mathbb{R}$ and $n$ is a positive integer.
The first problem given in this section is to find \[\lim_{t\rightarrow 2}\dfrac{4(t^2 -4)}{t-2}\]
Let's try using the limit laws to compute this. We have
\begin{align}
\displaystyle\lim_{t\rightarrow 2}\dfrac{4(t^2 -4)}{t-2} &= \dfrac{\displaystyle\lim_{t\rightarrow 2} 4(t^2 - 4)}{\displaystyle\lim_{t\rightarrow 2} (t-2)}
\\&= \dfrac{\displaystyle 4 (\lim_{t\rightarrow 2}t^2 - \lim_{t\rightarrow 2}4)}{\displaystyle\lim_{t\rightarrow 2} t-\lim_{t\rightarrow 2}2}
\\&= \dfrac{4(2^2 - 4)}{2 - 2}
\\&= \dfrac{0}{0}
\end{align}
The result is $0/0$, which we call an indeterminate form, which is different than dividing by zero.
To deal with these, we employ the following strategy:
Strategy for Evaluating Indeterminate Forms
Suppose you want to evaluate $\displaystyle \lim_{x\rightarrow a}f(x)$ but end up with $0/0$. Then:
Replace the original function $f(x)$ with a new function $g(x)$ that takes on the same values as the original function everywhere except at $x = a$. Usually:
The function $f$ has the limit $L$ as $x$ increases without bound (meaning $x\rightarrow \infty$), written as \[\lim_{x\rightarrow \infty}f(x) = L\] if $f(x)$ can be made arbitrarily close to $L$ by taking $x$ large enough.
Similarly, the function $f$ has the limit $M$ as $x$ decreases without bound (meaning $x\rightarrow -\infty$), written as \[\lim_{x\rightarrow -\infty}f(x) = M\] if $f(x)$ can be made arbitrarily close to $M$ by taking $x$ to be a sufficiently large negative number.
The same table and graph techniques apply for limits at infinity. Moreover, if $f(x)$ has a limit at infinity, then $f(x)$ has a horizontal asymptote. Witness:
If \[f(x) = \begin{cases} -1 & x < 0 \\ 1 & x\geq 0\end{cases} \qquad g(x) = \dfrac{1}{x^2}\]
Find $\displaystyle \lim_{x\rightarrow \infty} f(x)$ and $\displaystyle \lim_{x\rightarrow-\infty}f(x)$
Find $\displaystyle \lim_{x\rightarrow \infty} g(x)$ and $\displaystyle \lim_{x\rightarrow-\infty}g(x)$
We will now look at how to find the limit at infinity of a rational function. First we need two facts. Recall:
A polynomial function of degree $n$ is a function of the form \[f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0\qquad a_n \neq 0\]
where $a_n, a_{n-1},\dots, a_1, a_0 \in \mathbb{R}$.
A rational function is a function of the form \[r(x) = \dfrac{p(x)}{q(x)}\] where $p(x)$ and $q(x)$ are both polynomials.
For all $n > 0$, we have \[\lim_{x\rightarrow \infty} \dfrac{1}{x^n} = 0 \qquad \text{and} \qquad \lim_{x\rightarrow-\infty}\dfrac{1}{x^n} = 0\] if $\frac{1}{x^n}$ is defined.
If we are interested in a limit at infinity of a rational function, first find the largest power of $x$ in the denominator, let's say $d$. Then, divide each term in the numerator and denominator by $x^d$ and use the limit laws!