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Section 1.6 Limit Theorems

Subsection 1.6.1 Basic Limit Theorems

The topics covered in this text are built on many results which we will not take time to investigate. Determining limits from the \(\varepsilon\)-\(\delta\) definition is very time consuming and theorems to calculate limits make the process much quicker. The following theorem states that "well behaved" functions may be combined and the resulting limit will be just as "well behaved."

If \(f\) and \(g\) are two functions, \(a\) is a real number, and \(\lim_{x \to a} f(x)\) and \(\lim_{x \to a} g(x)\) exist, then the following equations hold:

  1. \(\displaystyle \lim_{x \to a} (f \pm g)(x) = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)\)
  2. \(\displaystyle \lim_{x \to a} (f \cdot g)(x) = \left( \lim_{x \to a} f(x) \right) \cdot \left(\lim_{x \to a} g(x)\right)\)
  3. \(\displaystyle \lim_{x \to a} \left( \dfrac{f}{g} \right) (x) = \dfrac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}\)
  4. \(\lim_{x \to a} (c \cdot f)(x) = c \cdot \lim_{x \to a} f(x)\) for any constant \(c.\)

These results state that functions made by combining two functions will have limits determined by the original functions.

For any number \(a\text{,}\) prove that \(\lim_{x \rightarrow a}x=a\text{.}\) Hence, by the limit theorems, we know \(\lim_{x \rightarrow a}3x^2=3a^2\text{.}\)

Solution

For any number \(\varepsilon \gt 0\text{,}\) we must find a number \(\delta \gt 0\) so that if \(0 \lt |x-a| \lt \delta\) then \(|x-a| \lt \varepsilon\text{.}\) We see that in this case, if \(\delta = \varepsilon\) then the definition of limit is satisfied. So, we have established the fact that \(\lim_{x \rightarrow a} x = a.\)

Now, define \(f(x) = x\text{.}\) By the limit theorem, the limit of the product of two functions is the product of the limits of the two functions. Hence,

\begin{equation*} \lim_{x \rightarrow a} x^2= \left( \lim_{x \rightarrow a} x \right) \cdot \left( \lim_{x \rightarrow a} x \right) = a \cdot a = a^2\text{.} \end{equation*}

Furthermore, we can multiply \(x^2\) by a constant and the limit as \(x\) approaches \(a\) will be the the limit times the constant:

\begin{equation*} \lim_{x \rightarrow a} 3x^2= 3 \cdot \left( \lim_{x \rightarrow a} x^2\right) = 3a^2\text{.} \end{equation*}
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Subsection 1.6.2 Some Special Limits

Rather than spend more time on building the underpinnings for calculating all limits, we will focus our attention in this section to proving two useful limits. We need one more limit theorem which we state without proof.

The Squeeze Theorem. Suppose \(f,\) \(g\) and \(h\) are functions where over an interval containing the number \(a\) and \(g(x) \leq f(x) \leq h(x) \) (except possibly at \(a\)). If

\begin{equation*} \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, \end{equation*}

then \(\lim_{x \to a} f(x) = L.\)

Prove that

(a) \(\lim_{x \rightarrow 0} \sin(x) = 0\) and

(b) \(\lim_{x \rightarrow 0} \cos(x) = 1\)

Solution

(a)

(b)

We can build on the previous example to show the sine and cosine functions are continuous.

Prove that for any real number \(c\text{:}\)

(a) \(\lim_{x \rightarrow c} \sin(x) = \sin(c)\) and

(b) \(\lim_{x \rightarrow c} \cos(x) = \cos(c)\)

Solution

(a) Note that \(\sin(x) = \sin(c+x-c)\) and as \(x\) gets close to \(c\text{,}\) we see that \(x-c\) becomes close to \(0\text{.}\) So we may rewrite the limit as

\begin{equation*} \lim_{x \rightarrow c} \sin(x) = \lim_{h \rightarrow 0} \sin(c+h). \end{equation*}

We use the sum formula from the sine to get,

\begin{equation*} \lim_{h \rightarrow 0} \sin(c+h) = \lim_{h \rightarrow 0} \sin(c)\cos(h) +\cos(c)\sin(h), \end{equation*}

The sum rule for limits gives

\begin{equation*} \lim_{h \rightarrow 0} \sin(c+h) = \lim_{h \rightarrow 0} \sin(c)\cos(h) + \lim_{h \rightarrow 0} \cos(c)\sin(h). \end{equation*}

Since \(c\) is a constant, \(\sin(c)\) and \(\cos(c)\) are constants and may be factored out of the limits by the limit rules and the limit becomes.

\begin{equation*} \lim_{h \rightarrow 0} \sin(c+h) = \sin(c) \cdot \lim_{h \rightarrow 0} \cos(h) +\cos(c) \cdot \lim_{h \rightarrow 0}\sin(h) \end{equation*}

Since \(\lim_{h \rightarrow 0} \sin(h) = 0\) and \(\lim_{h \rightarrow 0} \cos(h) = 1\)

\begin{equation*} \lim_{h \rightarrow 0} \sin(c+h) = \sin(c) \cdot 1 +\cos(c) \cdot 0 \end{equation*}

Hence,

\begin{equation*} \lim_{x \rightarrow c} \sin(x) = \sin(c) \end{equation*}

(b)

In the activity we will use the Squeeze Theorem to prove two special limits.

Subsection 1.6.3 Summary

  • We may calculate a limit for a function quickly if it can be decomposed into a sum/difference, product or quotient of functions with known limits.

  • We established the two "special" limits

    \begin{equation*} \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 \;\; \mbox{ and } \;\; \lim_{x \rightarrow 0} \frac{\cos(x)-1}{x} = 0 \end{equation*}

    where \(x\) is given in radians.

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