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Precise Definitions of Limits

Definition 1.1.1.  Let \( f \) be a function defined on some open interval that contains the number \( a \), except possibly at \( a \) itself. Then we say that the limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \), and we write

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

if for every number \( \varepsilon > 0 \) there is a number \( \delta > 0 \) such that

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad |\ f(x) - L\ | < \varepsilon \)

Remark 1.1.2.  Since \( |\ x - a\ | \) is the distance from \( x \) to \( a \) and \( |\ f(x) - L\ | \) is the distance from \( f(x) \) to \( L \), and since \( \varepsilon \) can be arbitrarily small, the definition of a limit can be expressed in words as follows:

\( \displaystyle \lim_{x \to a}f(x) = L \) means that the distance between \( f(x) \) and \( L \) can be made arbitrarily small by requiring that the distance from \( x \) to a be sufficiently small (but not \( 0 \)).

Alternatively,

\( \displaystyle \lim_{x \to a}f(x) = L \) means that the values of \( f(x) \) can be made as close as we please to \( L \) by requiring \( x \) to be close enough to \( a \) (but not equal to \( a \)).

Example 1.1.3.  Prove that \( \displaystyle \lim_{x \to 1}\frac{2 + 4x}{3} = 2 \).

Solution.  Given \( \varepsilon > 0 \), we need \( \delta > 0 \) such that if \( 0 < |\ x − 1\ | < \delta \), then \( \displaystyle \left|\ \frac{2 + 4x}{3} - 2\ \right| < \varepsilon \). But

\( \displaystyle \begin{align} \left|\ \frac{2 + 4x}{3} - 2\ \right| < \varepsilon \quad &\Leftrightarrow \quad \left|\ \frac{4x - 4}{3}\ \right| < \varepsilon \\[7pt] &\Leftrightarrow \quad \left|\ \frac{4}{3}\ \right| \left|\ x - 1\ \right| < \varepsilon \\[7pt] &\Leftrightarrow \quad \left|\ x - 1\ \right| < \frac{3}{4}\varepsilon \end{align} \)

So if we choose \( \displaystyle \delta = \frac{3}{4}\varepsilon \), then \( \displaystyle 0 < \left|\ x - 1\ \right| < \delta \quad \Rightarrow \quad \left|\ \frac{2 + 4x}{3} - 2\ \right| < \varepsilon \). Thus, \( \displaystyle \lim_{x \to 1}\frac{2 + 4x}{3} = 2 \) by Definition 1.1.1.

\( \Box \)

Definition 1.1.4.  Let \( f \) be a function defined on some open interval that contains the number \( a \), except possibly at \( a \) itself. Then we write

\( \displaystyle \lim_{x \to a^{-}}\! \! f(x) = L \)

if for every number \( \varepsilon > 0 \) there is a number \( \delta > 0 \) such that

\( \text{if} \quad a - \delta < x < a \quad \text{then} \quad |\ f(x) - L\ | < \varepsilon \)

and

\( \displaystyle \lim_{x \to a^{+}}\! \! f(x) = L \)

if for every number \( \varepsilon > 0 \) there is a number \( \delta > 0 \) such that

\( \text{if} \quad a < x < a + \delta \quad \text{then} \quad |\ f(x) - L\ | < \varepsilon \)

Definition 1.1.5.  Let \( f \) be a function defined on some open interval that contains the number \( a \), except possibly at \( a \) itself. Then

\( \displaystyle \lim_{x \to a} f(x) = \infty \)

means that for every positive number \( M \) there is a positive number \( \delta \) such that

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad f(x) > M \)

and

\( \displaystyle \lim_{x \to a} f(x) = -\infty \)

means that for every negative number \( N \) there is a positive number \( \delta \) such that

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad f(x) < N \)

Example 1.1.6.  Prove that \( \displaystyle \lim_{x \to -3}\!\frac{1}{(x + 3)^{4}} = \infty \).

Solution.  Given \( M > 0 \), we need \( \delta > 0 \) such that \( \displaystyle 0 < |\ x +3\ | < \varepsilon \ \Rightarrow \ \frac{1}{(x + 3)^{4}} > M \). Now

\( \displaystyle \begin{align} \frac{1}{(x + 3)^{4}} > M \quad &\Leftrightarrow \quad (x + 3)^4 < \frac{1}{M} \\[7pt] &\Leftrightarrow \quad |\ x + 3\ | < \frac{1}{\sqrt[4]{M}} \\[7pt] \end{align} \)

So take \( \displaystyle \delta = \frac{1}{\sqrt[4]{M}} \). Then \( \displaystyle 0 < |\ x + 3\ | < \delta = \frac{1}{\sqrt[4]{M}} \ \Rightarrow \ \frac{1}{(x + 3)^{4}} > M \), so \( \displaystyle \lim_{x \to -3}\frac{1}{(x + 3)^{4}} = \infty \).

\( \Box \)

Definition 1.1.7.  Let \( f \) be a function defined on some interval \( \left(a,\ \infty \right) \). Then

\( \displaystyle \lim_{x \to \infty}\! f(x) = L \)

means that for every \( \varepsilon > 0 \) there is a corresponding number \( M \) such that

\( \text{if} \quad x > M \quad \text{then} \quad |\ f(x) - L\ | < \varepsilon \)

and let \( f \) be a function defined on some interval \( \left(-\infty,\ a \right) \). Then

\( \displaystyle \lim_{x \to -\infty}\!\!\! f(x) = L \)

means that for every \( \varepsilon < 0 \) there is a corresponding number \( N \) such that

\( \text{if} \quad x < N \quad \text{then} \quad |\ f(x) - L\ | < \varepsilon \)

Definition 1.1.8.  Let \( f \) be a function defined on some interval \( \left(a,\ \infty \right) \). Then

\( \displaystyle \lim_{x \to \infty}\! f(x) = \infty \)

means that for every positive number \( M \) there is a corresponding positive number \( N \) such that

\( \text{if} \quad x > N \quad \text{then} \quad f(x) > M \)

Calculating Limits

Theorem 1.2.1.  Suppose that \( c \) is a constant and the limits

\( \displaystyle \lim_{x \to a}f(x) \quad \text{and} \quad \lim_{x \to a}g(x) \)

exist. Then

Proof.  Let \( \displaystyle \lim_{x \to a}f(x) = L \) and \( \displaystyle \lim_{x \to a}g(x) = M \).

(1) Let \( \varepsilon > 0 \) be given. We must find \( \delta > 0 \) such that

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad |\ f(x) + g(x) - ( L + M )\ | < \varepsilon \)

Using the Triangle Inequality we can write

\( \displaystyle \begin{align} |\ f(x) + g(x) - ( L + M )\ | &= |\ ( f(x) - L ) + ( g(x) - M )\ | \\[7pt] &\le |\ f(x) - L\ | + |\ g(x) - M\ | \end{align} \)

We make \( |\ f(x) + g(x) - ( L + M )\ | \) less than \( \varepsilon \) by making each of the terms \( |\ f(x) - L\ | \) and \( |\ g(x) - M\ | \) less than \( \varepsilon/2 \). Since \( \varepsilon/2 > 0 \) and \( \displaystyle \lim_{x \to a}f(x) = L \), there exists a number \( \delta_{1} > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta_{1} \quad \text{then} \quad |\ f(x) - L\ | < \frac{\varepsilon}{2} \)

Similarly, since \( \displaystyle \lim_{x \to a}g(x) = M \), there exists a number \( \delta_{2} > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta_{2} \quad \text{then} \quad |\ g(x) - M\ | < \frac{\varepsilon}{2} \)

Let \( \delta = \min\{\delta_{1},\delta_{2}\} \), the smaller of the numbers \( \delta_{1} \) and \( \delta_{2} \). Notice that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad 0 < |\ x - a\ | < \delta_{1} \quad \text{and} \quad 0 < |\ x - a\ | < \delta_{2} \)

and so

\( \displaystyle |\ f(x) - L\ | < \frac{\varepsilon}{2} \quad \text{and} \quad |\ g(x) - M\ | < \frac{\varepsilon}{2} \)

Therefore,

\( \displaystyle \begin{align} |\ f(x) + g(x) - ( L + M )\ | &\le |\ f(x) - L\ | + |\ g(x) - M\ | \\[7pt] &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align} \)

To summarize,

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad |\ f(x) - L\ | < \varepsilon \)

Thus, by the definition of a limit,

\( \displaystyle \lim_{x \to a}\left[ f(x) + g(x) \right] = L + M \)

\( \Box \)

(4) Let \( \varepsilon > 0 \) be given. We want to find \( \delta > 0 \) such that

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad |\ f(x)g(x) - LM\ | < \varepsilon \)

In order to get terms that contain \( |\ f(x) - L\ | \) and \( |\ g(x) - M\ | \), we add and subtract \( Lg(x) \) as follows:

\( \displaystyle \begin{align} |\ f(x)g(x) - LM\ | &= |\ f(x)g(x) - Lg(x) + Lg(x) - LM\ | \\[7pt] &= |\ \left[ f(x) - L \right]g(x) + L \left[ g(x) - M \right]\ | \\[7pt] &\le |\ \left[ f(x) - L \right]g(x)\ | + |\ L\left[ g(x) - M \right]\ | \\[7pt] &= |\ f(x) - L\ ||\ g(x)\ | + |\ L\ ||\ g(x) - M\ | \end{align} \)

We want to make each of these terms less than \( \varepsilon/2 \). Since \( \displaystyle \lim_{x \to a}g(x) = M \), there exists a number \( \delta_{1} > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta_{1} \quad \text{then} \quad |\ g(x) - M\ | < \frac{\varepsilon}{2(1 + |\ L\ |)} \)

Also, there is a number \( \delta_{2} > 0 \) such that if \( 0 < |\ x - a\ | < \delta_{2} \), then

\( \displaystyle |\ g(x) - M\ | < 1 \)

and therefore

\( \displaystyle |\ g(x)\ | = |\ g(x) - M + M\ | \le |\ g(x) - M\ | + |\ M\ | < 1 + |\ M\ | \)

Since \( \displaystyle \lim_{x \to a}f(x) = L \), there exists a number \( \delta_{3} > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta_{3} \quad \text{then} \quad |\ f(x) - L\ | < \frac{\varepsilon}{2(1 + |\ M\ |)} \)

Let \( \delta = \min\{\delta_{1},\delta_{2},\delta_{3} \} \). If \( 0 < |\ x - a\ | < \delta \), then we have \( 0 < |\ x - a\ | < \delta_{1} \), \( 0 < |\ x - a\ | < \delta_{2} \), and \( 0 < |\ x - a\ | < \delta_{3} \), so we can combine the inequalities to obtain

\( \displaystyle \begin{align} |\ f(x)g(x) - LM\ | &\le |\ f(x) - L\ ||\ g(x)\ | + |\ L\ ||\ g(x) - M\ | \\[7pt] &< \frac{\varepsilon}{2(1 + |\ M\ |)}(1 + |\ M\ |) + |\ L\ |\frac{\varepsilon}{2(1 + |\ L\ |)} \\[7pt] &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align} \)

This shows that

\( \displaystyle \lim_{x \to a}\left[ f(x)g(x) \right] = LM \)

\( \Box \)

(5) First let us show that

\( \displaystyle \lim_{x \to a}\frac{1}{g(x)} = \frac{1}{M} \)

To do this we must show that, given \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad \left|\ \frac{1}{g(x)} - \frac{1}{M}\ \right| < \varepsilon \)

Observe that

\( \displaystyle \left|\ \frac{1}{g(x)} - \frac{1}{M}\ \right| = \frac{|\ M - g(x)\ |}{|\ Mg(x)\ |} \)

We know that we can make the numerator small. But we also need to know that the denominator is not small when \( x \) is near \( a \). Since \( \displaystyle \lim_{x \to a}g(x) = M \), there is a number \( \delta_{1} > 0 \) such that, whenever \( 0 < |\ x - a\ | < \delta_{1} \), we have

\( \displaystyle |\ g(x) - M\ | < \frac{|\ M\ |}{2} \)

and therefore

\( \displaystyle \begin{align} |\ M\ | &= |\ M - g(x) + g(x)\ | \le |\ M - g(x)\ | + |\ g(x)\ | \\[7pt] &< \frac{|\ M\ |}{2} + |\ g(x)\ | \end{align} \)

This shows that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta_{1} \quad \text{then} \quad |\ g(x)\ | > \frac{|\ M\ |}{2} \)

and so, for these values of \( x \),

\( \displaystyle \frac{1}{|\ Mg(x)\ |} = \frac{1}{|\ M\ ||\ g(x)\ |} < \frac{1}{|\ M\ |} \cdot \frac{2}{|\ M\ |} = \frac{2}{M^{2}} \)

Also, there exists \( \delta_{2} > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta_{2} \quad \text{then} \quad |\ g(x) - M\ | < \frac{M^{2}}{2}\varepsilon \)

Let \( \delta = \min\{\delta_{1},\delta_{2} \} \). Then, for \( 0 < |\ x - a\ | < \delta \), we have

\( \displaystyle \left|\ \frac{1}{g(x)} - \frac{1}{M}\ \right| = \frac{|\ M - g(x)\ |}{|\ Mg(x)\ |} < \frac{2}{M^{2}} \frac{M^{2}}{2}\varepsilon = \varepsilon \)

It follows that \( \displaystyle \lim_{x \to a}\frac{1}{g(x)} = \frac{1}{M} \). Finally, using (4), we obtain

\( \displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\left[ f(x) \cdot \frac{1}{g(x)} \right] = \lim_{x \to a}f(x) \lim_{x \to a}\frac{1}{g(x)} = L \cdot \frac{1}{M} = \frac{L}{M} \)

\( \Box \)

Theorem 1.2.2.  If \( f(x) \le g(x) \) for all \( x \) in an open interval that contains \( a \) (except possibly at \( a \)) and

\( \displaystyle \lim_{x \to a} f(x) = L \quad \text{and} \quad \lim_{x \to a} g(x) = M \)

then \( L \le M \).

Proof.  Let \( \displaystyle \lim_{x \to a}f(x) = L \) and \( \displaystyle \lim_{x \to a}g(x) = M \). We use the method of proof by contradiction. Suppose, if possible, that \( L > M \). Theorem 1.2.1 - (2) says that

\( \displaystyle \lim_{x \to a}\left[ g(x) - f(x) \right] = M - L \)

Therefore, for any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that

\( \displaystyle \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad |\ \left[ g(x) - f(x) \right] - (M - L)\ | < \varepsilon \)

In particular, taking \( \varepsilon = L - M \) (noting that \( L - M > 0 \) by hypothesis), we have a number \( \delta > 0 \) such that

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad | \left[ g(x) - f(x) \right] - ( M - L ) | < L - M \)

Since \( b \le |b| \) for any number \( b \), we have

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad \left[ g(x) - f(x) \right] - ( M - L ) < L - M \)

which simplifies to

\( \text{if} \quad 0 < |\ x - a\ | < \delta \quad \text{then} \quad g(x) < f(x) \)

But this contradicts \( f(x) \le g(x) \). Thus the inequality \( L > M \) must be false. Therefore \( L \le M \).

\( \Box \)

Theorem 1.2.3. (The Squeeze Theorem)  If \( f(x) \le g(x) \le h(x) \) for all \( x \) in an open interval that contains \( a \) (except possibly at \( a \)) and

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

then

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

Proof.  Let \( \varepsilon > 0 \) be given. Since \( \displaystyle \lim_{x \to a}f(x) = L \), there is a number \( \delta_{1} > 0 \) such that

\( \text{if} \quad 0 < |\ x - a\ | < \delta_{1} \quad \text{then} \quad |\ f(x) - L\ | < \varepsilon \)

that is,

\( \text{if} \quad 0 < |\ x - a\ | < \delta_{1} \quad \text{then} \quad L - \varepsilon < f(x) < L + \varepsilon \)

Since \( \displaystyle \lim_{x \to a}h(x) = L \), there is a number \( \delta_{2} > 0 \) such that

\( \text{if} \quad 0 < |\ x - a\ | < \delta_{2} \quad \text{then} \quad |\ h(x) - L\ | < \varepsilon \)

that is,

\( \text{if} \quad 0 < |\ x - a\ | < \delta_{2} \quad \text{then} \quad L - \varepsilon < h(x) < L + \varepsilon \)

Let \( \delta = \min \{ \delta_{1}, \delta_{2} \} \). If \( 0 < |\ x - a\ | < \delta \), then \( 0 < |\ x - a\ | < \delta_{1} \) and \( 0 < |\ x - a\ | < \delta_{2} \), so

\( L - \varepsilon < f(x) \le g(x) \le h(x) < L + \varepsilon \)

In particular,

\( L - \varepsilon < g(x) < L + \varepsilon \)

and so \( |\ g(x) - L\ | < \varepsilon \). Therefore \( \displaystyle \lim_{x \to a}g(x) = L \).

\( \Box \)

Example 1.2.4.  Show that \( \displaystyle \lim_{x \to 0}x^{2} \sin \frac{1}{x} = 0 \)

Solution.  First note that we cannot calculate the limit directly because \( \displaystyle \lim_{x \to 0}\sin \frac{1}{x} \) does not exist. Instead we apply Theorem 1.2.3, and so we need to find a function \( f \) smaller than \( \displaystyle g(x) = x^{2} \sin \frac{1}{x} \) and a function \( h \) bigger than \( g \) such that both \( f(x) \) and \( h(x) \) approach \( 0 \). Because the sine of any number lies between \( -1 \) and \( 1 \), we can write

\( \displaystyle -1 \le \sin \frac{1}{x} \le 1 \)

Any inequality remains true when multiplied by a positive number. We know that \( x^{2} \ge 0 \) for all \( x \) and so, multiplying each side of the inequalities by \( x^{2} \), we get

\( \displaystyle -x^{2} \le x^{2} \sin \frac{1}{x} \le x^{2} \)

as illustrated by Figure 1.

Figure 1

We know that

\( \displaystyle \lim_{x \to 0}x^{2} = 0 \quad \text{and} \quad \lim_{x \to 0}(-x^{2}) = 0 \)

Taking \( f(x) = x^{2} \), \( \displaystyle g(x) = x^{2} \sin \frac{1}{x} \), and \( \displaystyle h(x) = x^{2} \) in the Theorem 1.2.3, we obtain

\( \displaystyle \lim_{x \to 0}x^{2} \sin \frac{1}{x} = 0 \)

\( \Box \)

The body text and the figures were excerpted from James Stewart, Calculus 8E.