Math 242: Calculus I

1.6 Properties of Limits

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1.6.1 Properties of Limits

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There are always three ideas associated with any new topic in this course.

1. The geometrical intuition

2. The difficult definition

3. The properties for performing computations

In the previous section we discussed the geometric intuition of convergence and the Weierstrauss definition of a limit of a function as the independent variable approaches a point in its domain.

We will continue to struggle with the geometric intuition. The geometry of the Cartesian plane is familiar to us. However the geometries of magnetic fields, Heisenberg condensate, and Non-newtonian fluids are more difficult to understand.

We will need to memorize and remember always the difficult definitions of our calculus because "The Devil is in the details." The difficult to understand details of calculus will reveal that seemly intuitively clear results are actually false. While results that appear completely wrong will turn out to be true.

Theorem 1.6.1

Suppose that point $a$ is an element of the domains of both function $f$ and $g$, $c$ is a real constant, and that the limits

$$ \displaystyle\lim_{x\rightarrow a} f(x) = L $$

and

$$ \displaystyle\lim_{x\rightarrow a} g(x) = M $$

both exist. Then the following limits also exist

$$ \begin{align*} 1.&\ \displaystyle\lim_{x\rightarrow a} (f+g)(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{f(x)}\right.} f(x) + g(x) \pmb{\left.\vphantom{g(x)}\right]} = \displaystyle\lim_{x\rightarrow a} f(x) + \displaystyle\lim_{x\rightarrow a} g(x) = L + M \\ \\ 2.&\ \displaystyle\lim_{x\rightarrow a} (f-g)(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{f(x)}\right.} f(x) - g(x) \pmb{\left.\vphantom{g(x)}\right]} = \displaystyle\lim_{x\rightarrow a} f(x) - \displaystyle\lim_{x\rightarrow a} g(x) = L - M \\ \\ 3.&\ \displaystyle\lim_{x\rightarrow a} (cf)(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{c}\right.} cf(x) \pmb{\left.\vphantom{f(x)}\right]} = c\,\displaystyle\lim_{x\rightarrow a} f(x) = cL \\ \\ 4.&\ \displaystyle\lim_{x\rightarrow a} \left( fg \right)(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{f(x)}\right.} f(x)g(x) \pmb{\left.\vphantom{g(x)}\right]} = \displaystyle\lim_{x\rightarrow a} f(x) \cdot \displaystyle\lim_{x\rightarrow a} g(x) = LM \\ \\ 5.&\ \displaystyle\lim_{x\rightarrow a} \left( \frac{f}{g} \right)(x) = \displaystyle\lim_{x\rightarrow a} \left[ \frac{f(x)}{g(x)} \right] = \frac{\displaystyle\lim_{x\rightarrow a} f(x)}{\displaystyle\lim_{x\rightarrow a} g(x)} = \frac{L}{M},\ \text{if }\displaystyle\lim_{x\rightarrow a} g(x) \neq 0 \end{align*} $$

Proof

Here we will prove some of these properties.

1. $\displaystyle\lim_{x\rightarrow a} (f+g)(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{f(x)}\right.} f(x) + g(x) \pmb{\left.\vphantom{g(x)}\right]} = \displaystyle\lim_{x\rightarrow a} f(x) + \displaystyle\lim_{x\rightarrow a} g(x) = L + M$

Given the hypotheses of theorem 1.6.1 and an $\epsilon > 0$, we know that the limits $\displaystyle\lim_{x\rightarrow a} f(x) = L$ and $\displaystyle\lim_{x\rightarrow a} g(x) = M$ already exist. Using the definition of the limit of a function we know we have

• $\delta_1>0$ so that $|x-a|<\delta_1$ implies that $|f(x)-L|<\frac{\epsilon}{2}$, and

• $\delta_2>0$ so that $|x-a|<\delta_2$ implies that $|g(x)-M|<\frac{\epsilon}{2}$.

Define $\delta = \min\left\{\delta_1,\ \delta_2\right\}$. Then whenever $|x-a|<\delta$, we have that

$$ | (f+g)(x) - (L + M) | = | f(x) - L + g(x) - M | \le | f(x) - L | + | g(x) - M | < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$

Therefore,

$$ \displaystyle\lim_{x\rightarrow a} (f+g)(x) = L+M $$

4. $\displaystyle\lim_{x\rightarrow a} \left( fg \right)(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{f(x)}\right.} f(x)g(x) \pmb{\left.\vphantom{g(x)}\right]} = \displaystyle\lim_{x\rightarrow a} f(x) \cdot \displaystyle\lim_{x\rightarrow a} g(x) = LM$

Step 1:

Given the hypotheses of theorem 1.6.1 and an $\epsilon > 0$, we know that the limits $\displaystyle\lim_{x\rightarrow a} f(x) = L$. Using the defintion of the limit of a function we know we have

• $\delta_1>0$ so that $|x-a|<\delta_1$ implies that $|f(x)-L|<\sqrt{\frac{\epsilon}{2}}$

• $\delta_2>0$ so that $|x-a|<\delta_2$ implies that $|f(x)-L|<\frac{\epsilon}{4|L|}$

Define $\delta = \min\left\{\delta_1,\ \delta_2\right\}$. Then whenever $|x-a|<\delta$, we have that

$$ \begin{align*} | (f^2)(x) - L^2 | &= \left| \left( f(x) - L \right)\left( f(x) + L\right) \right| \\ \\ &= \left| \left( f(x) - L \right)\left( f(x) - L + L + L \right) \right| \\ \\ &= \left| \left( f(x) - L \right)^2 + 2L\left( f(x) - L \right) \right| \\ \\ &\le \left| \left( f(x) - L \right)^2 \right| + \left| 2L\left( f(x) - L \right) \right| \\ \\ &= \left| \left( f(x) - L \right) \right|^2 + 2|L|\left| \left( f(x) - L \right) \right| \\ \\ &= \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align*} $$

Therefore,

$$ \displaystyle\lim_{x\rightarrow a} \left(f^2\right)(x) = L^2 $$

Step 2:

Consider the following equations

$$ \left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 = \frac{a^2 + 2ab + b^2}{4} - \frac{a^2 - 2ab + b^2}{4} = \frac{4ab}{4} = ab $$

This tells us that

$$ f(x)g(x) = \left( \frac{f(x) + g(x)}{2} \right)^2 - \left( \frac{f(x) - g(x)}{2} \right)^2 $$

• We know from property 1 that $\displaystyle\lim_{x\rightarrow a} f(x) + g(x) = L+M$
• We know from property 2 that $\displaystyle\lim_{x\rightarrow a} f(x) - g(x) = L-M$
• We know from property 3 that $\displaystyle\lim_{x\rightarrow a} \frac{f(x) + g(x)}{2} = \frac{L+M}{2}$
• We know from property 3 that $\displaystyle\lim_{x\rightarrow a} \frac{f(x) - g(x)}{2} = \frac{L-M}{2}$
• We know from step 1 that $\displaystyle\lim_{x\rightarrow a} \left( \frac{f(x) + g(x)}{2} \right)^2 = \left( \frac{L+M}{2} \right)^2$
• We know from step 1 that $\displaystyle\lim_{x\rightarrow a} \left( \frac{f(x) - g(x)}{2} \right)^2 = \left( \frac{L-M}{2} \right)^2$
• We know from property 2 that $\displaystyle\lim_{x\rightarrow a} \left( \frac{f(x) + g(x)}{2} \right)^2 - \left( \frac{f(x) - g(x)}{2} \right)^2 = \left( \frac{L+M}{2} \right)^2 - \left( \frac{L-M}{2} \right)^2$

Therefore,

$$ \displaystyle\lim_{x\rightarrow a} (fg)(x) = \left( \frac{L+M}{2} \right)^2 - \left( \frac{L-M}{2} \right)^2 = LM $$

$\tombstone$

Exercise 1

Compute $\displaystyle\lim_{x\rightarrow 3} \left( 4x^2 - 9x + 2 \right)$

Check Your Work

Recall from the previous section that $\displaystyle\lim_{x\rightarrow a} x = a$, so $\displaystyle\lim_{x\rightarrow 3} x = 3$. Therefore
$$ \displaystyle\lim_{x\rightarrow 3} \left( 4x^2 - 9x + 2 \right) = 4(3)^2 - 9(3) + 2 = 36 - 27 + 2 = 11 $$

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1.6.2 The Power Property

Theorem 1.6.2 The Power Property

Suppose that point $a$ is an element of the domain of function $f$, $n\in\mathbb{N}$ is a positive integer, and that the limit

$$ \displaystyle\lim_{x\rightarrow a} f(x) = L $$

exists. Then

$$ \begin{align*} 6.&\ \displaystyle\lim_{x\rightarrow a} \left( f^n \right)(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{f(x)^n}\right.} f(x)^n \pmb{\left.\vphantom{f(x)^n}\right]} = \left[ \displaystyle\lim_{x\rightarrow a} f(x) \right]^n = L^n \end{align*} $$

Proof:

This is a proof using the Principle of Mathematical Induction.

Step 1:

Given the hypotheses of theorem 1.6.2 and an $\epsilon > 0$, we know that the limit $\displaystyle\lim_{x\rightarrow a} f(x) = L$ exists. Thus we know that for the positive integer $k=1$,

$$ \displaystyle\lim_{x\rightarrow a} f^1(x) = \displaystyle\lim_{x\rightarrow a} f(x) = L = L^1 $$

Step 2:

Suppose that Theorem 1.6.2 is true for the first $n$ integers, $1,\ \dots,\ n$. We will prove that it must be true for $k=n+1$. Since $n\ge 1$, we know that $k = n+1 > 1$. This allows us to use property 4 of Theorem 1.6.1 as follows:

$$ \displaystyle\lim_{x\rightarrow a} f^{n+1}(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{f(x)^{n+1}}\right.} f(x)^{n+1} \pmb{\left.\vphantom{f(x)^{n+1}}\right]} = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{f(x)^n}\right.} f(x)^n\cdot f(x) \pmb{\left.\vphantom{f(x)}\right]} = \displaystyle\lim_{x\rightarrow a} f^n(x) \cdot \displaystyle\lim_{x\rightarrow a} f(x) $$

The induction hypothesis "Suppose that Theorem 1.6.2 is true for the first $n$ integers, $1,\ \dots,\ n$" means that we already know that

$$ \displaystyle\lim_{x\rightarrow a} f^n(x) = L^n $$

Therefore,

$$ \displaystyle\lim_{x\rightarrow a} f^{n+1}(x) = \displaystyle\lim_{x\rightarrow a} f^n(x) \cdot \displaystyle\lim_{x\rightarrow a} f(x) = L^n\cdot L = L^{n+1} $$

We just used the hypothesis that if the theorem is true for integer $n$, then it must also be true for integer $n+1$. In step 1 we proved that Theorem 1.6.2 is true for $n=1$.

• Since Theorem 1.6.1 is true for $n=1$, then by step 2, it is also true for $n=2$.
• Since Theorem 1.6.1 is true for $n=2$, then by step 2, it is also true for $n=3$.
• Since Theorem 1.6.1 is true for $n=3$, then by step 2, it is also true for $n=4$.
• Since Theorem 1.6.1 is true for $n=4$, then by step 2, it is also true for $n=5$.
• etc.

Step 3:

By the Principle of Mathematical Induction Theorem 1.6.2 is true for all positive integers.

This is always the third step! $\tombstone$

Exercise 2

Compute $\displaystyle\lim_{x\rightarrow 3} \left( 4x^2 - 9x + 2 \right)^3$

Check Your Work

$$ \displaystyle\lim_{x\rightarrow 3} \left( 4x^2 - 9x + 2 \right)^3 = \left( 4(3)^2 - 9(3) + 2 \right)^3 = \left( 36 - 27 + 2 \right)^3 = 11^3 = 121\cdot 11 = 1331. $$

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1.6.3 The Root Property

Theorem 1.6.3 The Root Property

Suppose that point $a$ is an element of the domain of function $f$, $n\in\mathbb{N}$ is a positive integer, and that the limit

$$ \displaystyle\lim_{x\rightarrow a} f(x) = L $$

exists. Then

$$ \begin{align*} 7.&\ \displaystyle\lim_{x\rightarrow a} \left( \sqrt[n]{f} \right)(x) = \displaystyle\lim_{x\rightarrow a} \pmb{\left[\vphantom{\sqrt[n]{f(x)}}\right.} \sqrt[n]{f(x)} \pmb{\left.\vphantom{\sqrt[n]{f(x)}}\right]} = \sqrt[n]{ \displaystyle\lim_{x\rightarrow a} f(x) } = \sqrt[n]{L} = L^{\frac{1}{n}} \end{align*} $$

Using these seven properties, we can compute limits of all of the algebraic functions. For example

If $c$ is a real constant and $n$ is a positive integer, then

\begin{align*} 8.&\ \displaystyle\lim_{x\rightarrow a} c = c \\ \\ 9.&\ \displaystyle\lim_{x\rightarrow a} x = a \\ \\ 10.&\ \displaystyle\lim_{x\rightarrow a} x^n = a^n \\ \\ 11.&\ \displaystyle\lim_{x\rightarrow a} \sqrt[n]{x} = \sqrt[n]{a} \\ \end{align*}

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1.6.4 One-Sided Limits

We defined one-sided limits in section 1.5. A one-sided limit computes the limit of a function results when we approach a value in the domain from only the right or only the left.

• The right-hand limit results as the independent variable decreases along the horizontal axis toward $a$ from the right.

• The left-hand limit results as the independent variable increases along the horizontal axis toward $a$ from the left.

Example 1

Figure 1

The Heaviside function $H(x) = \left\{ \begin{array}{rcl} x & \text{ if } & 1\ge 0 \\ \\ 0 & \text{ if } & x < 0 \end{array} \right.$

There a jump discontinuit at the $x=0$. A jump discontinuity occurs when the left-hand limit and the right-hand limit are different.

Right-hand limit

If we follow value of the output of the Heaviside function as we approach zero from the right we get that the output is always $1$.

$$ \begin{array}{cccccccccc} \hline x & 1 & .5 & .25 & .125 & .0625 & .03125 & .01 & .001 & .00001 \\ \hline y & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array} $$

We write

$$ \displaystyle\lim_{x\rightarrow 0^+} H(x) = \displaystyle\lim_{x\rightarrow 0^+} 1 = 1 $$

We place the plus sign in the exponent position of the value $x$ approaches to indicate that this is a one-sided limit from the right-hand side.

Left-hand limit

If we follow value of the output of the Heaviside function as we approach zero from the left we get that the output is always $0$.

$$ \begin{array}{cccccccccc} \hline x & -1 & -0.5 & -0.25 & -0.125 & -0.0625 & -0.03125 & -0.01 & -0.001 & -0.00001 \\ \hline y & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array} $$

We write

$$ \displaystyle\lim_{x\rightarrow 0^-} H(x) = \displaystyle\lim_{x\rightarrow 0^+} 0 = 0 $$

We place the minus sign in the exponent position of the value $x$ approaches to indicate that this is a one-sided limit from the left-hand side.

Exercise 3

Compute the left-hand and right-hand limits of $f$ as $x$ approaches $\frac{\pi}{2}$ where

$$ f(x) = \tan(x) $$

Solution

Figure 2

$$ \begin{array}{cccccc} \hline x & \frac{3\pi}{4} & \frac{5\pi}{8} & \frac{9\pi}{16} & \frac{17\pi}{32} & \frac{33\pi}{64} &\frac{65\pi}{128} \\ \hline y & -1 & -2.414 & -5.027 & -10.153 & -20.355 & -40.735 \\ \hline \end{array} $$
The limit as $x$ approaches $\frac{\pi}{2}$ from the right is
$$ \displaystyle\lim_{x\rightarrow\frac{\pi}{2}^+} \tan(x) = -\infty $$
$$ \begin{array}{cccccc} \hline x & \frac{1\pi}{4} & \frac{3\pi}{8} & \frac{7\pi}{16} & \frac{15\pi}{32} & \frac{31\pi}{64} &\frac{63\pi}{128} \\ \hline y & 1 & 2.414 & 5.027 & 10.153 & 20.355 & 40.735 \\ \hline \end{array} $$
The limit as $x$ approaches $\frac{\pi}{2}$ from the left is
$$ \displaystyle\lim_{x\rightarrow\frac{\pi}{2}^-} \tan(x) = +\infty $$

Theorem 1.6.4

The two-sided limit exits if and only if the one-sided limits both exist and they are equal.

$$ \displaystyle\lim_{x\rightarrow a} f(x) = L $$

if and only if

$$ \displaystyle\lim_{x\rightarrow a^-} f(x) = L = \displaystyle\lim_{x\rightarrow a^+} f(x) $$

Exercise 4

Compute the left-hand and right-hand limits of $f$ as $x$ approaches $-1$ where

$$ f(x) = \frac{x+1}{x^3+1} $$

What is the value of the two-sided limit

$$ \displaystyle\lim_{x\rightarrow -1} f(x) = ? $$

Solution

Figure 3

$$ \begin{align*} \displaystyle\lim_{x\rightarrow -1^+} f(x) &= \displaystyle\lim_{x\rightarrow -1^+} \frac{x+1}{x^3+1} = \displaystyle\lim_{x\rightarrow -1^+} \frac{x+1}{(x+1)(x^2-x+1)} = \displaystyle\lim_{x\rightarrow -1^+} \frac{1}{x^2 - x + 1} = \frac{1}{3} \\ \\ \displaystyle\lim_{x\rightarrow -1^-} f(x) &= \displaystyle\lim_{x\rightarrow -1^-} \frac{1}{x^2-x+1} = \frac{1}{3} \end{align*} $$
Since the left-hand limit and the right-hand limit are equal, the normal two-sided limit exists and
$$ \displaystyle\lim_{x\rightarrow -1} f(x) = \frac{1}{3} $$

Exercise 5

Compute the left-hand and right-hand limits of $g$ as $x$ approaches $0$ where

$$ g(x) = \sqrt{x} $$

What is the value of the two-sided limit

$$ \displaystyle\lim_{x\rightarrow 0} g(x) = ? $$

Solution

Figure 4

$$ \begin{align*} \displaystyle\lim_{x\rightarrow 0^+} g(x) &= \displaystyle\lim_{x\rightarrow 0^+} \sqrt{x} = \sqrt{\displaystyle\lim_{x\rightarrow 0^+} x} = \sqrt{0} = 0 \\ \\ \displaystyle\lim_{x\rightarrow 0^-} g(x) &\ {\color{red}\text{Does Not Exit}} \end{align*} $$
because there are no elements of the domain of $g$ to the left of $0$ on the number line. Since one of the one-sided limits does not exist, the two-sided limit does not exist
$$ \displaystyle\lim_{x\rightarrow 0} g(x) \ {\color{red}\text{Does Not Exit}} $$

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1.6.5 The Squeeze Theorem

Theorem 1.6.5 The Squeeze Theorem

If three functions all have a neighborhood of a point $a$, say $(a-h,a+h)$ is a neighbord in the domain of functions $f$, $g$, and $h$. Furthermore, if on this interval we have that for all $x\in(a-h,a+h)$, except possibly as $a$ itself, we have

$$ f(x) \le g(x) \le h(x) $$

And finally, we also know that both the lower function $f$ and the upper function $h$ have the same limit

$$ \displaystyle\lim_{x\rightarrow a} f(x) = L = \displaystyle\lim_{x\rightarrow a} h(x) $$

Then the limit of $g$ as $x$ approaches $a$ also exist and

$$ \displaystyle\lim_{x\rightarrow a} g(x) = L $$

Example 2

Compute the limit of the function $g$ as $x$ approaches 0, where

$$ g(x) = x^2\cos\left(\frac{1}{x}\right) $$

Now we already know that the absolute value of cosine is a number in the interval $[0,1]$ because the range of cosine is $[-1,1]$. We also know that the expression $x^2$ is always $\ge 0$.

This yields the inequality

$$ |g(x)|= \left|\,x^2\cos\left(\frac{1}{x}\right)\,\right| = \left|x^2\right|\,\left|\,\cos\left(\frac{1}{x}\right)\,\right| \le x^2\cdot 1 = x^2 $$

We can use the definition of the absolute value function to write our inequality

$$ -x^2 \le |g(x)| \le x^2 $$

At this point we set $f(x)=-x^2$, and $g(x)=x^2$. All three of these functions $f$, $g$, and $h$ are defined on the interval $(-1,1)$ which includes zero. Furthermore we know from the properties of limits that

$$ \begin{align*} \displaystyle\lim_{x\rightarrow 0} f(x) &= \displaystyle\lim_{x\rightarrow 0} -x^2 = -\left(\displaystyle\lim_{x\rightarrow 0} x \right)^2 = 0 \\ \\ \displaystyle\lim_{x\rightarrow 0} h(x) &= \displaystyle\lim_{x\rightarrow 0} x^2 = \left(\displaystyle\lim_{x\rightarrow 0} x \right)^2 = 0 \end{align*} $$

Therefore, using the squeeze theorem

$$ \displaystyle\lim_{x\rightarrow 0} g(x) = \displaystyle\lim_{x\rightarrow 0} x^2\cos\left(\frac{1}{x}\right) = 0 $$

Figure 5

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