Theorem 1.5.2. Limit of \(x\).
\(x\) goes where it goes.
Section 1.5 Basic Limit Forms
Rather than guess the value of a limit and then attempt to prove it using a definition, we will use the definitions to prove some basic limit forms and prove some arithmetic properties of limits. These will allow us to evaluate limits similar to the way we would evaluate a function. The goal is to solve a problem like
\begin{equation*}
\lim_{x \to 5} \frac{3x^2-7}{x+3}
\end{equation*}
by using the properties to break it up into known, basic forms.
We will then show then demonstrate that this method does not always work. This restriction is part of the theme that we cannot treat the infinite like the finite.
Subsection 1.5.1 Basic Limits
\begin{equation*}
\lim_{x \to a} x = a.
\end{equation*}
|
\begin{equation*}
\lim_{x \to \infty} x = \infty
\end{equation*}
|
\begin{equation*}
\lim_{x \to -\infty} x = -\infty
\end{equation*}
|
\begin{equation*}
\lim_{x \to a} k = k.
\end{equation*}
|
\begin{equation*}
\lim_{x \to \infty} k = k.
\end{equation*}
|
\begin{equation*}
\lim_{x \to 0^+} \frac{1}{x} = \infty.
\end{equation*}
|
\begin{equation*}
\lim_{x \to 0^-} \frac{1}{x} = -\infty.
\end{equation*}
|
\begin{equation*}
\lim_{x \to 0} \frac{1}{x^2} = \infty.
\end{equation*}
|
\begin{equation*}
\lim_{x \to \infty} \frac{1}{x} = 0.
\end{equation*}
|
Note any two sided limit implies both one sided limits. For example \(\lim_{x \to a^+} x = a.\) is also true. This implication says simply if the limit is defined from both sides, then it is certainly defined from one (either) side. Also the limit \(\lim_{x \to 0} \frac{1}{x^2} = \infty\) is actually not a basic form, but rather a result of the previous two in the list. However, it is convenient to memorize it at this point.
Proofs of some of these basic forms are provided for three reasons. First, who would want to base their calculations or solutions or statements that are only appear to be true because we looked at insufficient examples? Truth here is means that we can arrive at this conclusion based on basic facts (think axioms and definitions) and logic.
If you object that you are more than willing to accept these as true, trusting the mathematicians to not pull a fast one on you, then note the second reason to present these proofs is that they are part of mathematics culture. For many mathematicians “if there is no proof it is not mathematics.” For those who think, “if there is no application it is not meaningful,” you should understand that these are both statements of personal values.
Finally, remember that evaluating limits is not an important task in life, that is, it is unlikely you will do much if any of this in your careers. Some of these proofs in addition to establishing truth, help us make sense of the statement. For example what does it mean that \(1/x^2\) becomes infinite as \(x\) approaches 0? What does infinity even mean?
Proof.
\begin{align*}
\lim_{x \to a} x \amp = a.\\
|f(x)-L| \amp < \epsilon. \amp \amp \text{Start with the epsilon portion of the definition.} \\
|x-a| \amp < \epsilon. \amp \amp \text{Substitute} f(x)=x \text{ and } L=a.\\
|x-a| \amp < \delta. \amp \amp \text{End with the delta portion of the definition.}
\end{align*}
Thus for all \(\epsilon > 0\) there exists a \(\delta=\epsilon>0\) such that the definition is true. Note that no algebra at all was needed.
Theorem 1.5.3. Limit of a constant.
A constant is where it is.
Proof.
\begin{align*}
\lim_{x \to a} K \amp = K.\\
|f(x)-L| \amp < \epsilon. \amp \amp \text{Start with the epsilon portion of the definition.} \\
|K-K| \amp < \epsilon. \amp \amp \text{Substitute} f(x)=K \text{ and } L=K.\\
0 \amp < \epsilon. \amp \amp \text{This is true regardless of } \delta
\end{align*}
Thus for all \(\epsilon > 0\) there exists a \(\delta=1>0\) such that the definition is true.
This next limit helps us understand what infinity means in one context.
Theorem 1.5.4. Limit of \(1/x\) to 0 from the right.
In a limit, \(1/0=\infty\text{.}\)
Proof.
\begin{align*}
\lim_{x \to 0^+} \frac{1}{x} \amp = \infty.\\
f(x) \amp > N. \amp \amp \text{Start with the vertical portion of the definition.} \\
\frac{1}{x} \amp > N. \amp \amp \text{Substitute} f(x)=1/x.\\
1 \amp > xN. \amp \amp \text{Multiply both sides.}\\
\frac{1}{N} \amp > x. \amp \amp \text{Multiply both sides.}\\
\frac{1}{N} \amp > x-0. \amp \amp \text{Multiply both sides.}\\
x \amp < \delta. \amp \amp \text{End with the horizontal portion of the definition.}
\end{align*}
Thus for all \(N > 0\) there exists a \(\delta=\frac{1}{N}>0\) such that the definition is true. Note if \(N\) is a very large number then \(\delta\) will be a very small number.
Theorem 1.5.5. Limit of \(1/x\) to infinity.
In a limit, \(1/\infty=0\text{.}\)
Proof.
\begin{align*}
\lim_{x \to \infty} \frac{1}{x} \amp = 0.\\
|f(x)-L| \amp < \epsilon. \amp \amp \text{Start with the vertical portion of the definition.} \\
\left|\frac{1}{x}-0\right| \amp < \epsilon. \amp \amp \text{Substitute } f(x)=1/x \mbox{ and } L=0. \\
\frac{1}{x} \amp < \epsilon. \\
1 \amp < x\epsilon. \amp \amp \text{Multiply both sides. Note x is always positive.}\\
\frac{1}{\epsilon} \amp < x. \amp \amp \text{Multiply both sides. Note epsilon is always positive.} \\
x \amp > M. \amp \amp \text{End with the horizontal portion of the definition.}
\end{align*}
Thus for all \(\epsilon > 0\) there exists a \(M=\frac{1}{\epsilon}>0\) such that the definition is true. Note if epsilon is a small number than \(M\) will be a very large number.
Checkpoint 1.5.6.
Prove the following limit by modifying Theorem 1.5.4.
\begin{equation*}
\lim_{x \to 0^-} \frac{1}{x} = -\infty.
\end{equation*}
Theorem 1.5.7. Limit of \(1/x^2\) to infinity.
A square hides a multitude of evils.
Proof.
We know already that \(\lim_{x \to 0^+} \frac{1}{x} = \infty\) and \(\lim_{x \to 0^-} \frac{1}{x} = -\infty\text{.}\) Consider
\begin{align*}
\lim_{x \to 0^+} \frac{1}{x^2} \amp = \\
\lim_{x \to 0^+} \left(\frac{1}{x}\right)^2 \amp = \amp \text{algebra}\\
\left(\lim_{x \to 0^+} \frac{1}{x}\right)^2 \amp = \infty \text{ exponent property}
\end{align*}
Notice that these steps would not change if we approaches 0 from the left because the square would eliminate the negatives. Thus because the limit is defined and has the same result from the left and right then the limit (two sided) is also defined.
This last proof requires some ideas explained below.
Subsection 1.5.2 Properties of Limits
Now that we know the limits of some very basic forms, we need to learn how to combine these forms to evaluate more complicated forms.
Claim 1.5.8. The limit of a sum is the sum of the limits.
In symbols
\begin{equation*}
\lim_{x \to a} [f(x)+g(x)] = \left[ \lim_{x \to a} f(x) \right] + \left[ \lim_{x \to a} g(x) \right].
\end{equation*}
To understand why this is correct consider Figure 1.5.9. This shows the sum of \(f(x)\) (red line) and \(g(x)\) (blue line) as \(f(x)+g(x)\) (purple line). Before changing the point or epsilon, check the following. The box around \(f(x)\) with width \(2\delta_f\) and height \(\epsilon/2\) contains all of \(f(x)\text{.}\) Similarly the box around \(g(x)\) with width \(2\delta_g\) and height \(\epsilon/2\) contains all of \(g(x)\text{.}\) Finally check that the box around \(f(x)+g(x)\) with width \(\min(\delta_f,\delta_g)\) (which is \(\delta_g\) in this case) and height \(\epsilon\) contains all of \(f(x)+g(x)\text{.}\) Next try changing epsilon and the point to see that the relationship still holds.
Instructions.
Confirm that if the smaller of the two deltas for \(\epsilon/2\) is selected, the sum function is always within \(\epsilon\text{.}\) You can change the point and epsilon.
The limit of sum is sum of the limits.
The goal is to show that
\begin{equation*}
\lim_{x \to a} [f(x)+g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x).
\end{equation*}
Suppose \(\lim_{x \to a} f(x)=L_f\) and \(\lim_{x \to a} g(x)=L_g.\) By definition of these limits for all \(\epsilon/2\) there exists some \(\delta_f > 0\) and \(\delta_g > 0\) such that
\begin{equation*}
|x-a| < \delta_f \text{ implies } |f(x)-L_f| < \epsilon/2
\end{equation*}
and
\begin{equation*}
|x-a| < \delta_g \text{ implies } |g(x)-L_g| < \epsilon/2.
\end{equation*}
Thus for \(\delta=\min(\delta_f,\delta_g)\)
\begin{align*}
|x-a| \amp < \delta \amp \text{ implies } \amp \amp |f(x)-L_f| \amp < \epsilon/2\\
|x-a| \amp < \delta \amp \text{ implies } \amp \amp |g(x)-L_g| \amp < \epsilon/2\\
|x-a| \amp < \delta \amp \text{ implies } \amp \amp |f(x)-L_f|+|g(x)-L_g| \amp < \epsilon\\
|x-a| \amp < \delta \amp \text{ implies } \amp \amp |f(x)-L_f+g(x)-L_g| \amp < \epsilon\\
|x-a| \amp < \delta \amp \text{ implies } \amp \amp |[f(x)+g(x)]-[L_f+L_g]| \amp < \epsilon
\end{align*}
Thus for all \(\epsilon > 0\) there exists a \(\delta=\min(\delta_f,\delta_g) > 0\) such that \(|x-a| < \delta\) implies \(|[f(x)+g(x)]-[L_f+L_g]| < \epsilon\text{.}\)
Note that each of the properties in Table 1.5.10 require that the individual limits exist. The Limit of Composition property requires that \(f(x)\) be a particularly ‘nice’ function. The necessary property will be presented in a later section.
| Limit of Scalar Product |
\begin{equation*}
\lim_{x \to a} cf(x) = c \lim_{x \to a} f(x)
\end{equation*}
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| Limit of Sum |
\begin{equation*}
\lim_{x \to a} [ f(x)+g(x) ] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)
\end{equation*}
|
| Limit of Product |
\begin{equation*}
\lim_{x \to a} [ f(x)g(x) ] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)
\end{equation*}
|
| Limit of Quotient |
\begin{equation*}
\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}
\end{equation*}
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| Limit of Composition |
\begin{equation*}
\lim_{x \to a} f(g(x)) = f\left( \lim_{x \to a} g(x) \right)
\end{equation*}
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| Limit of Exponents |
\begin{equation*}
\lim_{x \to a} f(x)^r = \left( \lim_{x \to a} f(x) \right)^r
\end{equation*}
|
The exponent property is actually a special case of the composition property, but it is convenient for us to memorize it for use in this chapter.
Subsection 1.5.3 More Limit Forms
The following are also expected to be memorized for the sake of evaluation of limits. Each of these will match your intuition from the graphs of these functions.
\begin{equation*}
\lim_{x \to a} \cos(x) = \cos(a).
\end{equation*}
|
Works for all trig |
\begin{equation*}
\lim_{x \to \infty} \cos(x) \text{ undefined}
\end{equation*}
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True for all trig |
\begin{equation*}
\lim_{x \to a} e^x= e^a.
\end{equation*}
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Works for all bases |
\begin{equation*}
\lim_{x \to \infty} e^x= \infty.
\end{equation*}
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Works for all bases |
\begin{equation*}
\lim_{x \to -\infty} e^x= 0.
\end{equation*}
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Works for all bases |
\begin{equation*}
\lim_{x \to a} \ln(x)= \ln(a).
\end{equation*}
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Works for all bases |
\begin{equation*}
\lim_{x \to \infty} \ln(x)= \infty.
\end{equation*}
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Works for all bases |
\begin{equation*}
\lim_{x \to 0^+} \ln(x)= -\infty.
\end{equation*}
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Works for all bases |
