Calculus BC · Unit 1
If a function is trapped between two functions that share a limit, it has that same limit.
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The Squeeze Theorem
If \(g(x) \leq f(x) \leq h(x)\) near \(x = a\) (except maybe at \(a\)), and
\(\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L\)
then \(\lim_{x \to a} f(x) = L\).
Both bounds must meet
The lower function and the upper function must have the same limit. If they go to different numbers, the theorem does not apply.
A bouncing factor
Sine and cosine stay between −1 and 1. Multiply by something that goes to 0, and the product is squeezed to 0.
\(-1 \leq \sin\theta \leq 1\)
A famous limit
\(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\)
This is proved with the Squeeze Theorem. You will use it often in Calculus BC.
Two polynomial or absolute-value bounds with the same limit.
\(g(x) = 4 - x^2\)
\(h(x) = 4 + x^2\)
\(\lim_{x \to 0} g(x) = 4\)
\(\lim_{x \to 0} h(x) = 4\)
The bounds have the same limit, so the Squeeze Theorem applies.
\(\lim_{x \to 0} f(x) = 4\)
Answer: \(4\)
\(\lim_{x \to 0}(3 - |x|) = 3\)
\(\lim_{x \to 0}(3 + |x|) = 3\)
Squeeze Theorem.
\(\lim_{x \to 0} f(x) = 3\)
Answer: \(3\)
\(\lim_{x \to 0} 1 = 1\)
\(\lim_{x \to 0}(1 + x^2) = 1\)
Squeeze Theorem.
\(\lim_{x \to 0} f(x) = 1\)
Answer: \(1\)
\(\lim_{x \to 0}(-x^2) = 0\)
\(\lim_{x \to 0} x^2 = 0\)
Squeeze Theorem.
\(\lim_{x \to 0} f(x) = 0\)
Answer: \(0\)
\(\lim_{x \to 2} 5 = 5\)
\(\lim_{x \to 2}\bigl(5 + (x-2)^2\bigr) = 5\)
Squeeze Theorem.
\(\lim_{x \to 2} f(x) = 5\)
Answer: \(5\)
\(\lim_{x \to 1}\bigl(7 - (x-1)^2\bigr) = 7\)
\(\lim_{x \to 1} 7 = 7\)
Squeeze Theorem.
\(\lim_{x \to 1} f(x) = 7\)
Answer: \(7\)
A sine or cosine that stays between −1 and 1, times something that goes to 0.
\(-1 \leq \sin\dfrac{1}{x} \leq 1\) for \(x \neq 0\)
Multiply through by \(x^2\) (which is ≥ 0).
\(-x^2 \leq x^2 \sin\dfrac{1}{x} \leq x^2\)
\(\lim_{x \to 0}(-x^2) = 0\)
\(\lim_{x \to 0} x^2 = 0\)
Squeeze Theorem.
\(\lim_{x \to 0} x^2 \sin\dfrac{1}{x} = 0\)
Answer: \(0\)
\(-|x| \leq x \sin\dfrac{1}{x} \leq |x|\)
\(\lim_{x \to 0}(-|x|) = 0\)
\(\lim_{x \to 0}|x| = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
\(-x^2 \leq x^2 \cos\dfrac{1}{x} \leq x^2\)
\(\lim_{x \to 0}(\pm x^2) = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
\(-|x| \leq |x|\cos\dfrac{1}{x^2} \leq |x|\)
\(\lim_{x \to 0}(\pm |x|) = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
\(-(x-1)^2 \leq (x-1)^2 \sin\dfrac{1}{x-1} \leq (x-1)^2\)
\(\lim_{x \to 1}(\pm (x-1)^2) = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
\(-|x|^3 \leq x^3 \sin\dfrac{1}{x} \leq |x|^3\)
\(\lim_{x \to 0}(\pm |x|^3) = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
A trig bound that is not just −1 to 1, or a squeeze as x goes to infinity.
For \(0 < x < \dfrac{\pi}{2}\),
\(\cos x \leq \dfrac{\sin x}{x} \leq 1\)
\(\lim_{x \to 0^+} \cos x = 1\)
\(\lim_{x \to 0^+} 1 = 1\)
Squeeze Theorem from the right: the right-hand limit is 1.
\(\dfrac{\sin x}{x}\) is even, so the left-hand limit is also 1.
\(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\)
Answer: \(1\)
\(-1 \leq \sin x \leq 1\)
\(-\dfrac{1}{x} \leq \dfrac{\sin x}{x} \leq \dfrac{1}{x}\) for \(x > 0\)
\(\lim_{x \to \infty}\left(-\dfrac{1}{x}\right) = 0\)
\(\lim_{x \to \infty}\dfrac{1}{x} = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
\(-1 \leq \cos\dfrac{1}{x} \leq 1\)
\(1 \leq 3 + 2\cos\dfrac{1}{x} \leq 5\)
\(x^2 \leq x^2\bigl(3 + 2\cos\dfrac{1}{x}\bigr) \leq 5x^2\)
\(\lim_{x \to 0} x^2 = 0\)
\(\lim_{x \to 0} 5x^2 = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
\(-\sqrt{x} \leq \sqrt{x}\,\sin\dfrac{1}{\sqrt{x}} \leq \sqrt{x}\) for \(x > 0\)
\(\lim_{x \to 0^+}(\pm\sqrt{x}) = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
\(-x^4 \leq x^4\sin\dfrac{1}{x} \leq x^4\)
\(x^2 - x^4 \leq x^2 + x^4\sin\dfrac{1}{x} \leq x^2 + x^4\)
\(\lim_{x \to 0}(x^2 - x^4) = 0\)
\(\lim_{x \to 0}(x^2 + x^4) = 0\)
Squeeze Theorem.
\(0\)
Answer: \(0\)
\(-x^2 \leq x^2\cos\dfrac{1}{x} \leq x^2\)
\(2 - x^2 \leq x^2\cos\dfrac{1}{x} + 2 \leq 2 + x^2\)
\(\lim_{x \to 0}(2 - x^2) = 2\)
\(\lim_{x \to 0}(2 + x^2) = 2\)
Squeeze Theorem.
\(2\)
Answer: \(2\)
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