Module 5: Limit Properties #

② We’ll try to get \(|g(x)||f(x)-L| < \frac{\varepsilon}{2}\).

We’ll use a similar idea as in ①, but \(|f(x)-L| < \frac{\varepsilon}{2(|g(x)|+1)}\) won’t work as this changes based on \(x\) (so \(\delta_\varepsilon\) would as well).

Instead: we will want \(|g(x)| \leq M_g\), some positive constant.

If we get this, then:

Positive, so we get \(\delta_f = \delta_\varepsilon\left( \frac{\varepsilon}{2M_g} \right)\).

Notice: As \(1 > 0\), we can find a \(\delta_1 > 0\) such that if \(|x-a|<\delta_1\), then \(||g(x)|-|M||<1\).

In particular, if \(|x-a|<\delta_1\), then \(|g(x)| < |M|+1 = M_g\)

(We will need to prove that if \(\lim_{x\to a} g(x) = M\), then \(\lim_{x\to a} |g(x)| = |M|\)).

(For now, we will assume this fact.)

“Idea” so far: We will three upper limits for \(\delta\):

This will give \(|g(x)| < |M| + 1\)

This will give \(|g(x)| |f(x) - L| < \frac{\varepsilon}{2} \)

This will give \(|L||g(x) - M| < \frac{\varepsilon}{2} \)

Putting this together: We let \(\delta = \min(\delta_1, \delta_f, \delta_g)\). For \(|x-a| < \delta\):

As \(|x-a| < \delta_1\), we have \(| |g(x)| - |M| | < 1\), so \(-1 < |g(x)| - |M| < 1\), and \(|M| - 1 < |g(x)| < |M| + 1\), so \(|g(x)| < |M| + 1\).

As \(|x-a| < \delta_f\), we have \(|f(x) - L| < \frac{\varepsilon}{2(|M|+1)}\), so \((|M|+1) |f(x) - L| < \frac{\varepsilon}{2}\).

As \(|g(x)| < |M| + 1\), \(|g(x)| |f(x) - L| < (|M|+1) |f(x) - L|\). Hence, \(|g(x)||f(x) - L| < \frac{\varepsilon}{2}\)

As \(|x-a| < \delta_g\), we have \(|g(x) - M| < \frac{\varepsilon}{2(|L|+1)}\), so \((|L|+1) |g(x) - M| < \frac{\varepsilon}{2}\).

As \(|L| < |L|+1\), \(|L||g(x) - M| < (|L|+1)|g(x) - M|\). Hence, \(|L||g(x) - M| < \frac{\varepsilon}{2}\).

If \(\lim_{x \to a} g(x) = M\), then \(\lim_{x \to a} |g(x)| = |M|\).

\text{If } |x-a| < \delta, \text{ then } | l(x) - l | < 3

Putting this together: \(|g(x)f(x) - LM| = |g(x)f(x) - Lg(x) + Lg(x) - LM| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon\)

As \(|g(x)f(x) - LM| = |g(x)f(x) - Lg(x) + Lg(x) - LM| \leq |g(x)(f(x) - L)| + |L(g(x) - M)|\) \ (by the triangle inequality), we have: \ \(|g(x)f(x) - Lg(x) + Lg(x) - LM| \leq |g(x)(f(x) - L)| + |L(g(x) - M)| < \epsilon\) \ \(|g(x)f(x) - Lg(x) + Lg(x) - LM| < \epsilon\) \ \(|f(x)g(x) - LM| < \epsilon\) \ \(|(fg)(x) - LM| < \epsilon\)

For the proof: Set-up/introduce D, E, f and g. Introduce a, L, M. Assume \(\lim_{x \to a} f(x) = L\) and \(\lim_{x \to a} g(x) = M\). Let \(\epsilon > 0\) be arbitrary.

• Use these assumed limits to define \(\delta_f\) (for \(\frac{\epsilon}{2(|M| + 1)}\)), \(\delta_g\) (for \(\frac{\epsilon}{2(|L| + 1)}\)), and \(\delta_1\) (for 1).
• Set \(\delta = \min(\delta_f, \delta_g, \delta_1)\) and explain why \(\delta > 0\).
• Let \(x \in D \setminus {a}\) be arbitrary and suppose \(|x - a| < \delta\).
• Show \(|(fg)(x) - LM| < \epsilon\). (Using the above arguments)
• Conclude the proof (building back up to quantified statement).

Proof: Let \(D \subseteq \mathbb{R}\), \(E \subseteq \mathbb{R}\). Let \(g : D \to E\) be a function and let \(a \in \mathbb{R}\) be an accumulation point of \(D\). Also, let \(M \in \mathbb{R}\) and suppose \(\lim_{x \to a} g(x) = M\).

We will prove that \(\lim_{x \to a} |g(x)| = |M|\).

Let \(\varepsilon > 0\) be arbitrary. As \(\lim_{x \to a} g(x) = M\), there is a \(\hat{\delta} > 0\) such that for any \(x \in D \setminus {a}\), if \(|x - a| < \hat{\delta}\), then \(|g(x) - M| < \varepsilon\).

We will consider three cases.

Case 1: Suppose \(M = 0\). Then, let \(\delta = \hat{\delta}\). As \(\hat{\delta} > 0\), \(\delta > 0\). Let \(x \in D \setminus {a}\) be arbitrary, and suppose \(|x - a| < \delta\). Then, we have that \(|x - a| < \hat{\delta}\) so that \(|g(x) - M| < \varepsilon\). In this case: \(|g(x) - 0| < \varepsilon\), so \(|g(x)| < \varepsilon\), and \(||g(x)|| < \varepsilon\). From this, \(||g(x)| - 0| < \varepsilon\), and \(||g(x)| - |0|| < \varepsilon\), so that \(| |g(x)| - |M| | < \varepsilon\). Hence, if \(|x - a| < \delta\), then \(||g(x)| - |M|| < \varepsilon\).

As \(x \in D \setminus {a}\) was arbitrary, for any \(x \in D \setminus {a}\), if \(|x - a| < \delta\), then \(||g(x)| - |M|| < \varepsilon\). As \(\delta > 0\), there exists \(\delta > 0\) such that for all \(x \in D \setminus {a}\), if \(|x - a| < \delta\), then \(||g(x)| - |M|| < \varepsilon\).

Case 2: Suppose \(M > 0\). Note that \(\frac{M}{2} > 0\) so that as \(\lim_{x\to a} g(x) = M\), there is a \(\hat{\delta} > 0\) such that for any \(x \in D \setminus {a}\), if \(|x-a| < \hat{\delta}\), then \(|g(x) - M| < \frac{M}{2}\). Let \(\delta = \min{\hat{\delta}, \tilde{\delta}}\). As \(\tilde{\delta} > 0\) and as \(\hat{\delta} > 0\), we have \(\delta > 0\). Now, let \(x \in D \setminus {a}\) be arbitrary, and suppose \(|x-a| < \delta\). Then \(|x-a| < \hat{\delta}\) so that:

As \(\frac{M}{2} < g(x)\) and \(\frac{M}{2} > 0\), we have that \(g(x) > 0\), so that \(|g(x)| = g(x)\).
As \(M > 0\), we have that \(|M| = M\).

Now, as \(|x-a| < \tilde{\delta}\), we have that \(|x-a| < \tilde{\delta}\) so that \(||g(x)| - |M|| < \epsilon\).

Hence, \(||g(x)| - |M|| < \epsilon\). Thus, if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \epsilon\). As \(x \in D \setminus {a}\) was arbitrary, for any \(x \in D \setminus {a}\), if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \epsilon\). As \(\tilde{\delta} > 0\), there exists \(\delta > 0\) such that for all \(x \in D \setminus {a}\), if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \epsilon\).

Proof: Let \(D \subseteq \mathbb{R}, E \subseteq \mathbb{R}\). Let \(g: D \rightarrow E\) be a function and let \(a \in \mathbb{R}\) be an accumulation point of \(D\). Also, let \(M \in \mathbb{R}\) and suppose \(\lim_{x \rightarrow a} g(x) = M\).

We will prove that \(\lim_{x \rightarrow a} |g(x)| = |M|\).

Let \(\epsilon > 0\) be arbitrary. As \(\lim_{x \rightarrow a} g(x) = M\), there is a \(\hat{\delta} > 0\) such that for any \(x \in D \setminus {a}\), if \(|x - a| < \hat{\delta}\), then \(|g(x) - M| < \epsilon\).

We will consider three cases.

Case 1: Suppose \(M = 0\). Then, let \(\delta = \hat{\delta}\). As \(\hat{\delta} > 0\), \(\delta > 0\). Let \(x \in D \setminus {a}\) be arbitrary, and suppose \(|x - a| < \delta\). Then, we have that \(|x - a| < \hat{\delta}\) so that \(|g(x) - M| < \epsilon\). In this case: \(|g(x) - 0| < \epsilon\), so \(|g(x)| < \epsilon\), and \(||g(x)|| < \epsilon\). From this, \(||g(x)| - |0|| < \epsilon\), and \(||g(x)| - 0|| < \epsilon\), so that \(||g(x)| - |M|| < \epsilon\). Hence, if \(|x - a| < \delta\), then \(||g(x)| - |M|| < \epsilon\).

As \(x \in D \setminus {a}\) was arbitrary, for any \(x \in D \setminus {a}\), if \(|x - a| < \delta\), then \(||g(x)| - |M|| < \epsilon\). As \(\delta > 0\), there exists \(\delta > 0\) such that for all \(x \in D \setminus {a}\), if \(|x - a| < \delta\), then \(||g(x)| - |M|| < \epsilon\).

Case 2: Suppose \(M > 0\). Note that \(\frac{M}{2} > 0\) so that as \(\lim_{x \to a} g(x) = M\), there is a \(\hat{\delta} > 0\) such that for any \(x \in D \setminus {a}\), if \(|x-a| < \hat{\delta}\), then \(|g(x) - M| < \frac{M}{2}\). Let \(\delta = \min(\hat{\delta}, \tilde{\delta})\). As \(\tilde{\delta} > 0\) and as \(\hat{\delta} > 0\), we have \(\delta > 0\). Now, let \(x \in D \setminus {a}\) be arbitrary, and suppose \(|x-a| < \delta\). Then \(|x-a| < \hat{\delta}\) so that:

As \(\frac{M}{2} < g(x)\) and \(\frac{M}{2} > 0\), we have that \(g(x) > 0\), so that \(|g(x)| = g(x)\). As \(M > 0\), we have that \(|M| = M\).

Now, as \(|x-a| < \tilde{\delta}\), we have that \(|x-a| < \tilde{\delta}\) so that \(|g(x)| - |M| < \epsilon\).

Hence, \(||g(x)| - |M|| < \epsilon\). Thus, if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \epsilon\). As \(x \in D \setminus {a}\) was arbitrary, for any \(x \in D \setminus {a}\), if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \epsilon\). As \(\delta > 0\), there exists \(\delta > 0\) such that for all \(x \in D \setminus {a}\), if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \epsilon\).

Case 3: Suppose that \(M < 0\). Note that \(\frac{|M|}{2} > 0\) so that as \(\lim_{x \to a} g(x) = M\), there is a \(\hat{\delta} > 0\) such that for any \(x \in D \backslash {a}\), if \(|x - a| < \hat{\delta}\), then \(|g(x) - M| < \frac{|M|}{2}\). Let \(\delta = \min(\hat{\delta}, \tilde{\delta})\). As \(\hat{\delta} > 0\) and \(\tilde{\delta} > 0\), we have \(\delta > 0\). Now, let \(x \in D \backslash {a}\) be arbitrary, and suppose \(|x-a| < \delta\). Then, we have that \(|x-a| < \hat{\delta}\) so that:

As \(M < 0\), \(|M| = -M\), so that:

Thus, \(g(x) < \frac{M}{2}\). As \(M < 0\), \(\frac{M}{2} < 0\) so that \(g(x) < 0\). Thus, \(|g(x)| = -g(x)\).

Now, as \(|x-a| < \delta\), we have \(|x-a| < \tilde{\delta}\) so that \(|g(x) - M| < \varepsilon\). Notice:

Hence, \(||g(x)| - |M|| < \varepsilon\). Thus, if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \varepsilon\). As \(x \in D \backslash {a}\) was arbitrary for any \(x \in D \backslash {a}\), if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \varepsilon\). As \(\delta > 0\), there exists \(\delta > 0\) such that for all \(x \in D \backslash {a}\), if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \varepsilon\).

These cases are exhaustive. Hence, there exists a \(\delta > 0\) such that for all \(x \in D \setminus {a}\), if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \epsilon\). As \(\epsilon > 0\) was arbitrary, for all \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x \in D \setminus {a}\), if \(|x-a| < \delta\), then \(||g(x)| - |M|| < \epsilon\). Therefore, \(\lim_{x \to a} |g(x)| = |M|\).

As \(D, E, g, a\) and \(M\) were arbitrary, for any \(D \subseteq \mathbb{R}, E \subseteq \mathbb{R}\) and any function \(g: D \to E\) and any accumulation point \(a \in \mathbb{R}\) of \(D\) and any \(M \in \mathbb{R}\), if \(\lim_{x \to a} g(x) = M\), then \(\lim_{x \to a} |g(x)| = |M|\).

The Triangle Inequality: For all real numbers \(x\) and \(y\), \(|x+y| \leq |x| + |y|\).

Proof: Let \(x\) and \(y\) be arbitrary real numbers. We will consider four cases.

Case 1: Suppose \(x < 0\) and \(y < 0\). Then, \(|x| = -x\) and \(|y| = -y\). Also, \(x+y < 0\), so \(|x+y| = -(x+y)\). Now:

\(|x+y| = -(x+y) = -x - y = (-x) + (-y) = |x| + |y|\).

Hence, \(|x+y| = |x| + |y|\), so that \(|x+y| \leq |x| + |y|\).

Case 2: Suppose \(x < 0\) and \(y \geq 0\). Then, \(|x| = -x\) and \(|y| = y\). We will consider two subcases.

Subcase 2.1: Suppose \(x+y \geq 0\). Then, \(|x+y| = x+y\). Now:

\(|x+y| = x+y \leq 2|x| + x + y = -2x + x + y = -x+y = |x| + |y|\).

Subcase 2.2: Suppose \(x+y < 0\). Then \(|x+y| = -(x+y)\). Now:

\(|x+y| = -(x+y) = -x -y \leq -x-y + 2|y| = -x-y+2y=-x+y = |x|+|y|\).

These subcases are exhaustive, so we have that \(|x+y| \leq |x| + |y|\).

Case 3: Suppose \(x \geq 0\) and \(y < 0\). Notice that \(|x+y| = |y+x|\), so by Case 2, we have \(|y+x| \leq |y| + |x| = |x| + |y|\). Hence, \(|x+y| \leq |x| + |y|\).

a = b = c < d = e = f \ a < f \ a < f

Case 4: Suppose \(x \geq 0\) and \(y \geq 0\). Then \(|x| = x\) and \(|y| = y\). Also \(x+y \geq 0\) so that \(|x+y| = x+y\). Now:

Thus, \(|x+y| = |x|+|y|\), so that \(|x+y| \leq |x|+|y|\). These cases are exhaustive, hence \(|x+y| \leq |x|+|y|\). As \(x\) and \(y\) were arbitrary, for all real numbers \(x\) and \(y\), \(|x+y| \leq |x|+|y|\).
The Reverse Triangle Inequality: For all real numbers \(x\) and \(y\), we have that:

Proof: Let \(x\) and \(y\) be arbitrary real numbers. We will consider four cases.

Case 1: Suppose \(x < 0\) and \(y < 0\). Then, \(|x| = -x\) and \(|y| = -y\). Now:

Hence, we have \(||x| - |y|| = |x-y|\), so that \(||x| - |y|| \leq |x-y|\).

Case 2: Suppose \(x < 0\) and \(y > 0\). Then, \(|x| = -x\) and \(|y| = y\). Now:

By the triangle inequality, \(|x+y| \leq |x|+|y|\), so we have:

Hence, we have \(||x|-|y|| \leq |x-y|\).

Case 3: Suppose \(x \geq 0\) and \(y < 0\). Then, note that by case 2, \(| |y| - |x| | \leq |y-x|\). Also, we have that: \(| |x| - |y| | = | -(|x| - |y|) | = | |y| - |x| |\), and also that: \(|x-y| = |-(x-y)| = |y-x|\). Hence, we have that \(||x|-|y|| \leq |x-y|\).

Case 4: Suppose \(x \geq 0\) and \(y \geq 0\). Then \(|x| = x\) and \(|y| = y\). Now: \(||x|-|y|| = |x-y|\). Thus, \(||x|-|y|| \leq |x-y|\).

These cases are exhaustive, hence: \(||x|-|y|| \leq |x-y|\). As \(x\) and \(y\) were arbitrary, for all real numbers \(x\) and \(y\), \(||x|-|y|| \leq |x-y|\).

By the Reverse Triangle Inequality:

Comment: The Reverse Triangle Inequality greatly simplifies the proof that

As \(||g(x)| - |M|| \leq |g(x) - M|\), when \(|g(x) - M| < \varepsilon\), we also have that \(||g(x)| - |M|| < \varepsilon\).

Definition/Notation: Suppose we have \(D \subseteq \mathbb{R}\) and \(E \subseteq \mathbb{R}\) and functions

We define a function \(\left(\frac{f}{g}\right): D \setminus {x \in D ,|, g(x) = 0} \to E\) by:

Our next ``goal" will be to show that: \(\displaystyle \lim_{x\to a} \left(\frac{f}{g}\right)(x) = \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\displaystyle\lim_{x\to a} f(x)}{\displaystyle\lim_{x\to a} g(x)} = \frac{L}{M}\)

Note: In order for this to make sense, we will need \(\displaystyle \lim_{x \to a} g(x) = M \neq 0\) (can’t divide by zero).

As we have \( \lim_{x \to a} f(x) = L \) and \( \lim_{x \to a} g(x) = M \), we try to get expressions of the form \(|f(x) - L|\) and \(|g(x) - M|\).

We already have \(M \neq 0\), but we’ll need \(|g(x)| \neq 0\) on some \(\delta\)-window around \(a\). As \(\lim_{x \to a} g(x) \neq 0\), this will be possible.