Using Limit Properties

Section 1.6 Using Limit Properties

Having proved the values of some simple limits and proved some limit properties, we will now combine these, occasionally with some creativity, to evaluate more limits. Most important are the reasons we need the creativity. This is part of the theme that we cannot treat the infinite like the finite and is part of the standard: Distinguish between meaningful calculations involving infinity and non-determinate ones.

Subsection 1.6.1 Using Basic Limit Properties

We have been able to prove the values of some simple limits 1.5.1 by guessing their value and using the definitions. We are able to prove some algebraic properties 1.5.10. Now we will demonstrate using those known limit forms and algebraic limit properties to determine the value of other slightly less simple limits. As you read the first example note how this differs from evaluating a function. Also note the role of order of operations.

Example 1.6.1. Evaluate a simple limit using properties.

\begin{equation*} \lim_{x \rightarrow 2} \frac{3x^2+2}{x^3-x} \end{equation*}

Solution.

\begin{align*} \lim_{x \rightarrow 2} \frac{3x^2+2}{x^3-x} = \amp \text{ using the quotient property} \\ \frac{\lim_{x \rightarrow 2} 3x^2+2}{\lim_{x \rightarrow 2} x^3-x} = \amp \text{ using the summation property} \\ \frac{\lim_{x \rightarrow 2} 3x^2+ \lim_{x \rightarrow 2} 2}{\lim_{x \rightarrow 2} x^3 + \lim_{x \rightarrow 2} -x} = \amp \text{ using the scalar property} \\ \frac{3\lim_{x \rightarrow 2} x^2+ \lim_{x \rightarrow 2} 2}{\lim_{x \rightarrow 2} x^3 - \lim_{x \rightarrow 2} x} = \amp \text{ using the exponent property} \\ \frac{3\left(\lim_{x \rightarrow 2} x\right)^2+ \lim_{x \rightarrow 2} 2}{\left(\lim_{x \rightarrow 2} x\right)^3 - \lim_{x \rightarrow 2} x} = \amp \text{ known limits} \\ \frac{3(2)^2+2}{2^3-2} = \frac{7}{3}. \end{align*}

Checkpoint 1.6.2.

(a)

What determined the order in which the algebraic limit properties were used? If you are unsure think about the order you would use to evaluate \(f(x)=\frac{3x^2+2}{x^3-x}\text{.}\)

(b)

How many properties were used at each step?

Example 1.6.3. Beyond Order of Operations.

\begin{equation*} \lim_{x \to 0} \frac{3x^2+2}{x^2} \end{equation*}

Solution.

\begin{align*} \lim_{x \to 0} \frac{3x^2+2}{x^2} = \amp \text{ using algebra}\\ \lim_{x \to 0} \frac{1}{x^2} \cdot (3x^2+2) = \amp \text{ using multiplication property} \\ \lim_{x \to 0} \frac{1}{x^2} \cdot \lim_{x \to 0} (3x^2+2) = \amp \text{ using summation property} \\ \lim_{x \to 0} \frac{1}{x^2} \cdot \left(\lim_{x \to 0} 3x^2+\lim_{x \to 0} 2\right) = \amp \text{ using scalar multiplication} \\ \lim_{x \to 0} \frac{1}{x^2} \cdot \left(3\lim_{x \to 0} x^2+\lim_{x \to 0} 2\right) = \amp \text{ using exponent property} \\ \lim_{x \to 0} \frac{1}{x^2} \cdot \left(3 \left[\lim_{x \to 0} x\right]^2+\lim_{x \to 0} 2\right) = \amp \infty \text{ known forms} \end{align*}

Checkpoint 1.6.4.

Compare the Example 1.6.1 and Example 1.6.3. How was the division handled differently? Why was it handled differently?

Note the last step in Example 1.6.3 requires proving that some ‘arithmetic’ we performed is reasonable. Next we look at examples that illustrate which arithmetic makes sense and which does not.

Example 1.6.5.

\(\lim_{x \to \infty} (x^2+x)(x+1)\)

Solution.

\begin{align*} \lim_{x \to \infty} (x^2+x)(x+1) = \amp\\ \lim_{x \to \infty} (x^2+x)\lim_{x \to \infty} (x+1) = \amp\\ \left(\lim_{x \to \infty} x^2+ \lim_{x \to \infty} x\right) \left(\lim_{x \to \infty} x+\lim_{x \to \infty} 1 \right) = \amp\\ \left(\left[\lim_{x \to \infty} x\right]^2+ \lim_{x \to \infty} x \right) \left(\lim_{x \to \infty} x+\lim_{x \to \infty} 1 \right) = \amp \end{align*}

The last line has three limits that are infinite. If they were finite we could use the limit properties to perform the arithmetic. Unfortunately, we cannot treat the infinite like the finite. We need additional theorems.

Theorem 1.6.6.

Under the following conditions each of the propositions below are true. \(\lim_{x \to \infty} f_1(x)=\infty\text{,}\) \(\lim_{x \to \infty} f_2(x)=\infty\text{,}\) \(\lim_{x \to \infty} g_p(x)=k\) where \(k\) is any real number, \(\lim_{x \to \infty} g_p(x)=k\) where \(k \gt 0\text{,}\) and \(\lim_{x \to \infty} g_n(x)=k\) where \(k \lt 0\text{.}\)

\(\lim_{x \to \infty} f_1(x)+f_2(x)=\infty\text{.}\)
\(\lim_{x \to \infty} f_1(x) \cdot f_2(x)=\infty\text{.}\)
\(\lim_{x \to \infty} f_1(x)+g(x)=\infty\text{.}\)
\(\lim_{x \to \infty} f_1(x) \cdot g_p(x)=\infty\text{.}\)
\(\lim_{x \to \infty} f_1(x) \cdot g_n(x)=-\infty\text{.}\)
\(\lim_{x \to \infty} f_1(f_2(x))=\infty\text{.}\)

Example 1.6.7.

We complete this example using the new theorem. \(\lim_{x \to \infty} (x^2+x)(x+1)\)

Solution.

\begin{align*} \lim_{x \to \infty} (x^2+x)(x+1) = \amp\\ \left(\left[\lim_{x \to \infty} x\right]^2+ \lim_{x \to \infty} x \right) \left(\lim_{x \to \infty} x+\lim_{x \to \infty} 1 \right) = \amp\\ \left(\lim_{x \to \infty} x^2+ \lim_{x \to \infty} x \right) \left(\lim_{x \to \infty} x+\lim_{x \to \infty} 1 \right) = \amp \text{ by 2} \\ \left(\lim_{x \to \infty} x^2+x \right) \left(\lim_{x \to \infty} x+\lim_{x \to \infty} 1 \right) = \amp \text{ by 1} \\ \left(\lim_{x \to \infty} x^2+x \right) \left(\lim_{x \to \infty} x+1 \right) = \amp \text{ by 3} \\ \lim_{x \to \infty} \left( x^2+x \right) \left( x+1 \right) = \amp \infty \text{ by 2} \end{align*}

Note that these theorems are also true if the limits are approaching a fininte number, e.g., \(\lim{x \to a} f(x)=\infty\text{.}\) Note as well that there are no theorems for subtraction and division. This is because they are problematic in a way that is first illustrated below and will be addressed in detail in Section 2.7.

Activity 1.6.1.

Why is the difference of two infinities not simply zero? We will address this by considering what would happen if we assumed it. Note the following examples use completely illegitimate notation: think of this as scratching out vague ideas in our head.

Guido supposed that \(\infty-\infty = 0\text{.}\)

\begin{align*} \lim_{x \to \infty} x^3-x^2+x & = \\ \infty - \infty + \infty & = \\ (\infty - \infty) + \infty & = \\ 0 + \infty & = \infty \end{align*}

Meanwhile Wolfgang supposed the same thing.

\begin{align*} \lim_{x \to \infty} x^3-x^2+x & = \\ \infty - \infty + \infty & = \\ \infty - (\infty + \infty) & = \\ \infty - \infty & = 0 \end{align*}

(a)

What did Guido and Wolfgang do differently?

(b)

Did they end up with the same result?

(c)

Is this a problem?

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