Limit of a Function
Let \( f \) be a function defined on a set \( S \), and let \( a \) be a limit point of \( S \). We say that \[ \lim_{x \to a} f(x) = L \] if for every \( \epsilon \gt 0 \), there exists a \( \delta \gt 0 \) such that for all \( x \in S \): \[ 0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \epsilon \]
Note: The condition \( 0 \lt |x - a| \) means we do not care about the value of \( f(a) \), or even if \( f(a) \) is defined. We only care about \( x \) near \( a \).
Visualizing the Epsilon-Delta Definition
To prove a limit exists, for any "target window" around \( L \) (height \( 2\epsilon \)), you must find a "launch window" around \( a \) (width \( 2\delta \)) such that every point in the launch window lands in the target window.
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Sequential Criterion (Theorem 20.2)
The limit \( \lim_{x \to a} f(x) = L \) exists if and only if:
This theorem allows us to use all the tools we developed for sequences (like the Squeeze Theorem, algebraic limit theorems) to prove things about functions.
Practice Problems
True/False
1. If \( \lim_{x \to a} f(x) = L \), then \( f(a) \) must equal \( L \).
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FALSE - The limit describes the behavior of \( f(x) \) as \( x \) approaches \( a \), not the value at \( a \). \( f(a) \) might not even be defined.
2. If \( \lim_{x \to a} f(x) = L \), then for every sequence \( (x_n) \) in the domain converging to \( a \) (with \( x_n \neq a \)), the sequence \( (f(x_n)) \) converges to \( L \).
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TRUE - This is the Sequential Criterion for limits.
3. If \( \lim_{x \to a} f(x) = L \) and \( \lim_{x \to a} g(x) = M \), then \( \lim_{x \to a} (f(x)g(x)) = LM \).
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TRUE - This is one of the algebraic limit theorems.
Fill in the Blank
4. To prove \( \lim_{x \to a} f(x) = L \) using the definition, for every \( \epsilon \gt 0 \), we must find a \( \delta \gt 0 \) such that if \( 0 \lt |x - a| \lt \delta \), then _____.
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\( |f(x) - L| \lt \epsilon \)
5. The Sequential Criterion states that \( \lim_{x \to a} f(x) = L \) if and only if for every sequence \( (x_n) \) in the domain with \( x_n \neq a \) and \( \lim x_n = a \), we have _____.
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\( \lim f(x_n) = L \)
6. If \( \lim_{x \to a} f(x) \gt 0 \), then there exists a neighborhood of \( a \) where \( f(x) \) is _____.
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positive
Full Problems
7. Use the \( \epsilon-\delta \) definition to prove that \( \lim_{x \to 2} (3x - 1) = 5 \).
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Let \( \epsilon \gt 0 \). We want to find \( \delta \gt 0 \) such that if \( 0 \lt |x - 2| \lt \delta \), then \( |(3x - 1) - 5| \lt \epsilon \).
Consider \( |(3x - 1) - 5| = |3x - 6| = 3|x - 2| \).
We want \( 3|x - 2| \lt \epsilon \), which is equivalent to \( |x - 2| \lt \epsilon/3 \).
Choose \( \delta = \epsilon/3 \). Then if \( 0 \lt |x - 2| \lt \delta \), we have \( |(3x - 1) - 5| = 3|x - 2| \lt 3(\epsilon/3) = \epsilon \).
8. Use the Sequential Criterion to show that \( \lim_{x \to 0} \sin(1/x) \) does not exist.
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Consider two sequences converging to 0:
\( x_n = \frac{1}{2\pi n} \). Then \( \sin(1/x_n) = \sin(2\pi n) = 0 \to 0 \).
\( y_n = \frac{1}{2\pi n + \pi/2} \). Then \( \sin(1/y_n) = \sin(2\pi n + \pi/2) = 1 \to 1 \).
Since the limits are different, the limit of the function does not exist.
9. Let \( f(x) = x^2 \sin(1/x) \) for \( x \neq 0 \). Show that \( \lim_{x \to 0} f(x) = 0 \).
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We know that \( -1 \le \sin(1/x) \le 1 \).
Multiplying by \( x^2 \) (which is non-negative), we get \( -x^2 \le x^2 \sin(1/x) \le x^2 \).
Since \( \lim_{x \to 0} (-x^2) = 0 \) and \( \lim_{x \to 0} x^2 = 0 \), by the Squeeze Theorem, \( \lim_{x \to 0} x^2 \sin(1/x) = 0 \).
10. Prove that if \( \lim_{x \to a} f(x) = L \) and \( \lim_{x \to a} f(x) = M \), then \( L = M \).
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Suppose \( L \neq M \). Let \( \epsilon = \frac{|L - M|}{2} \gt 0 \).
There exists \( \delta_1 \gt 0 \) such that \( 0 \lt |x - a| \lt \delta_1 \implies |f(x) - L| \lt \epsilon \).
There exists \( \delta_2 \gt 0 \) such that \( 0 \lt |x - a| \lt \delta_2 \implies |f(x) - M| \lt \epsilon \).
Let \( \delta = \min(\delta_1, \delta_2) \). For \( 0 \lt |x - a| \lt \delta \):
\( |L - M| = |L - f(x) + f(x) - M| \le |L - f(x)| + |f(x) - M| \lt \epsilon + \epsilon = 2\epsilon = |L - M| \).
This implies \( |L - M| \lt |L - M| \), a contradiction. Thus \( L = M \).