2.4: Limits: Algebraic Methods

The Limit Laws

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 $c$ is a constant and the limits \[\lim_{x\rightarrow a}f(x), \qquad \lim_{x\rightarrow a} g(x)\] exist. Then:
  1. $\displaystyle\lim_{x\rightarrow a} [f(x) + g(x)] = \lim_{x\rightarrow a} f(x) + \lim_{x\rightarrow a} g(x)$
  2. $\displaystyle\lim_{x\rightarrow a} [f(x) - g(x)] = \lim_{x\rightarrow a} f(x) - \lim_{x\rightarrow a} g(x)$
  3. $\displaystyle\lim_{x\rightarrow a} c\cdot f(x) = c\lim_{x \rightarrow a} f(x)$
  4. $\displaystyle\lim_{x \rightarrow a}[f(x)\cdot g(x)] = \left[\lim_{x\rightarrow a}f(x)\right]\cdot\left[\lim_{x\rightarrow a} g(x)\right]$
  5. $\displaystyle\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \dfrac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a}g(x)}$ if $\displaystyle\lim_{x\rightarrow a}g(x) \neq 0$.

We have seen these before!

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.
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

Consider \[\lim_{t\rightarrow 2}\dfrac{4(t^2 -4)}{t-2}\] 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 $x = a$. Then before applying limit laws, either:
    1. Cancel out common factors.
    2. Rationalize the numerator or denominator.
    3. Simplify the expression.
  2. Then find $\displaystyle\lim_{x\rightarrow a} g(x)$ instead.
Find $\lim_{x\rightarrow 1}g(x)$ where \[\begin{cases}x + 1 & x \neq 1 \\ \pi & x = 1\end{cases}\]
Find \[\lim_{x\rightarrow 2}\dfrac{4(x^2-4)}{x-2}\]
Find \[\lim_{h\rightarrow 0}\dfrac{\sqrt{1 + h} - 1}{h}\]
Find \[\lim_{h\rightarrow 0}\dfrac{(3+h)^2 - 9}{h}\]

Additional Properties of Limits

Recall:

\[\lim_{x\rightarrow a^+}f(x) = L = \lim_{x\rightarrow a^-}f(x) \qquad \text{if and only if} \qquad \lim_{x\rightarrow a}f(x) = L\]

Limit laws also hold for one-sided limits. This means instead of using the table technique, we can now check both left- and right-handed limits are equal with algebra!

Prove \[\lim_{x\rightarrow 0}\lvert x \rvert = 0\]
Prove $\displaystyle\lim_{x\rightarrow 0}\lvert x \rvert = 0$ does not exist.

Sometimes limits can't be computed. But if we trap the function within two other functions that we know the limit of, we can find the limit of the original function itself.

The Squeeze Theorem
If $f(x) \leq g(x) \leq h(x)$ when $x$ is near $a$ (except possibly at $a$) and we know \[\lim_{x\rightarrow a}f(x) = \lim_{x\rightarrow a}h(x) = L\] then $\displaystyle \lim_{x\rightarrow a}g(x) = L$.
Show that $\displaystyle \lim_{x\rightarrow 0} x^2 \sin \dfrac{1}{x} = 0$.

Limits of Trigonometric Functions

Here are a few limits to know.

\[\lim_{\theta \rightarrow a}\sin\theta = \sin a \qquad \lim_{\theta \rightarrow a}\cos\theta = \cos a \qquad \lim_{\theta \rightarrow 0} \dfrac{\sin\theta}{\theta} = 1\]

Notice for both $\sin \theta$ and $\cos \theta$ their limit values actually see what happens at $a$. The next section will describe why.

The core idea for trigonometric limits is to rewrite the expression in terms of limits you know how to evaluate.

Find $\displaystyle\lim_{x\rightarrow0} \dfrac{\sin 7x}{4x}$
Find $\displaystyle\lim_{\theta\rightarrow 0} \dfrac{\cos\theta - 1}{\theta}$