2.4: Limits

A First Look

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 $x$:
  1. $f(x) = \begin{cases} -1 & x < 0 \\ 1 & x \geq 0\end{cases} \ $ at $x = 0$.
  2. $\displaystyle g(x) = \frac{1}{x^2}$ 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
  1. $\displaystyle\lim_{x\rightarrow a} [f(x)]^r = \left[\lim_{x \rightarrow a}f(x)\right]^r = L^r$
  2. $\displaystyle\lim_{x\rightarrow a} c\cdot f(x) = c\lim_{x \rightarrow a} f(x) = cL$
  3. $\displaystyle\lim_{x\rightarrow a} [f(x) \pm g(x)] = \lim_{x\rightarrow a} f(x) \pm \lim_{x\rightarrow a} g(x) = L \pm M$
  4. $\displaystyle\lim_{x \rightarrow a}[f(x)\cdot g(x)] = \left[\lim_{x\rightarrow a}f(x)\right]\cdot\left[\lim_{x\rightarrow a} g(x)\right] = L\cdot M$
  5. $\displaystyle\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \dfrac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a}g(x)} = \frac{L}{M}$ where $M \neq 0$.
Given these functions Find
  1. $\displaystyle \lim_{x \rightarrow -2} [f(x) + 5g(x)]$
  2. $\displaystyle \lim_{x \rightarrow 1} [f(x) \cdot g(x)]$
  3. $\displaystyle \lim_{x \rightarrow 2} \dfrac{f(x)}{g(x)}$

  1. 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.
  1. $\displaystyle\lim_{x\rightarrow a} c = c$
  2. $\displaystyle\lim_{x\rightarrow a} x = a$
  3. $\displaystyle\lim_{x\rightarrow a} x^n = a^n$
  4. $\displaystyle\lim_{x \rightarrow a}\sqrt[n]{x} = \sqrt[n]{a}$
  5. If $f(x)$ is any function, then $\displaystyle\lim_{x\rightarrow a} \sqrt[n]{f(x)}= \sqrt[n]{\lim_{x \rightarrow a} f(x)}$.
Note: this says for $n$th roots and powers of $x$, we are allowed to substitute $a$ directly. We will see why this is true in Section 2.5.
Using the previous fact, find the following limits.
  1. $\displaystyle \lim_{x \rightarrow 2} x^3$
  2. $\displaystyle \lim_{x \rightarrow 4} 5x^{2}$
  3. $\displaystyle \lim_{x \rightarrow 1} (5x^4 - 2)$
  4. $\displaystyle \lim_{x \rightarrow 1} 2x^3\cdot \sqrt{x^2 + 7}$
  5. $\displaystyle \lim_{x \rightarrow 1} \dfrac{2x^2 + 1}{x + 1}$

First Look at Indeterminate Forms

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:
  1. 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:
    1. Cancel out common factors.
    2. Rationalize the numerator or denominator.
  2. Find $\displaystyle\lim_{x\rightarrow a} g(x)$ instead.
Find \[\lim_{x\rightarrow 2}\dfrac{4(x^2-4)}{x-2}\]
Find \[\lim_{h\rightarrow 0}\dfrac{\sqrt{1 + h} - 1}{h}\]

Limits at Infinity

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:

Find \[\lim_{x\rightarrow \infty} \dfrac{2x^2}{1 + x^2}\]
If \[f(x) = \begin{cases} -1 & x < 0 \\ 1 & x\geq 0\end{cases} \qquad g(x) = \dfrac{1}{x^2}\]
  1. Find $\displaystyle \lim_{x\rightarrow \infty} f(x)$ and $\displaystyle \lim_{x\rightarrow-\infty}f(x)$
  2. 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!

Find \[\lim_{x\rightarrow \infty}\dfrac{x^2 -x + 3}{2x^3 + 1}\]
If \[f(x) = \dfrac{2x^3 - 3x^2 + 1}{x^2 + 2x + 4}\] find $\displaystyle\lim_{x\rightarrow \infty}f(x)$ and $\displaystyle\lim_{x\rightarrow -\infty}f(x)$.