Calculus BC ยท Unit 1
Rewrite so a sine or tangent sits over the same angle, then the leftover constant is the limit.
Jump to section:
The famous limit
\(\lim_{\theta \to 0} \dfrac{\sin \theta}{\theta} = 1\)
Also \(\lim_{\theta \to 0} \dfrac{\tan \theta}{\theta} = 1\).
Match the inside
Rewrite so the same expression is in the sine and in the denominator. Then a leftover constant multiplies 1.
\(\dfrac{\sin 5x}{x} = 5 \cdot \dfrac{\sin 5x}{5x}\)
Rewrite tan and sec
\(\tan \theta = \dfrac{\sin \theta}{\cos \theta}\) and \(\sec \theta = \dfrac{1}{\cos \theta}\). Cosine goes to 1 as the angle goes to 0.
Do not substitute 0/0
Plugging in \(x = 0\) in \(\dfrac{\sin x}{x}\) gives 0/0. Use the famous limit instead.
Rewrite to \(\sin(kx)/(kx)\), then multiply by the leftover constant.
\(\dfrac{\sin 4x}{x} = 4 \cdot \dfrac{\sin 4x}{4x}\)
\(\lim_{x \to 0} \dfrac{\sin 4x}{4x} = 1\)
\(4 \cdot 1\)
\(4\)
Answer: \(4\)
\(\lim_{\theta \to 0} \dfrac{\sin \theta}{\theta} = 1\) with \(\theta = 3x\).
\(1\)
Answer: \(1\)
\(6 \cdot \dfrac{\sin 6x}{6x}\)
\(6 \cdot 1\)
\(6\)
Answer: \(6\)
\(\dfrac{2}{5} \cdot \dfrac{\sin 2x}{2x}\)
\(\dfrac{2}{5} \cdot 1\)
\(\dfrac{2}{5}\)
Answer: \(\dfrac{2}{5}\)
\(\dfrac{5}{10} \cdot \dfrac{\sin 5x}{5x}\)
\(\dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
\(\lim_{\theta \to 0} \dfrac{\tan \theta}{\theta} = 1\)
\(1\)
Answer: \(1\)
Tangent, or a constant that is not the same as the angle.
\(\tan 3x = \dfrac{\sin 3x}{\cos 3x}\)
\(\dfrac{\tan 3x}{x} = 3 \cdot \dfrac{\sin 3x}{3x} \cdot \dfrac{1}{\cos 3x}\)
\(3 \cdot 1 \cdot \dfrac{1}{1}\)
\(3\)
Answer: \(3\)
\(\dfrac{4}{8} \cdot \dfrac{\tan 4x}{4x}\)
\(\dfrac{1}{2} \cdot 1\)
\(\dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
\(\dfrac{7}{14} \cdot \dfrac{\sin 7x}{7x}\)
\(\dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
\(\lim_{x \to 0}(1 - \cos x) = 0\) and the denominator goes to 0, but \(1 - \cos x\) is even and the known limit is \(\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}\).
\(\dfrac{1 - \cos x}{x} = x \cdot \dfrac{1 - \cos x}{x^2}\)
\(0 \cdot \dfrac{1}{2}\)
\(0\)
Answer: \(0\)
\(\dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}\)
\(1 \cdot 1\)
\(1\)
Answer: \(1\)
\(\dfrac{5}{2} \cdot \dfrac{\tan 5x}{5x}\)
\(\dfrac{5}{2}\)
Answer: \(\dfrac{5}{2}\)
Rewrite tan and sec, then split into famous limits.
\(\tan 2x = \dfrac{\sin 2x}{\cos 2x}\)
\(\sec 5x = \dfrac{1}{\cos 5x}\)
\(\dfrac{\tan 2x}{4x \sec 5x} = \dfrac{\sin 2x}{\cos 2x} \cdot \dfrac{\cos 5x}{4x}\)
\(\dfrac{\sin 2x}{2x} \cdot \dfrac{2}{4} \cdot \dfrac{\cos 5x}{\cos 2x}\)
\(1 \cdot \dfrac{1}{2} \cdot \dfrac{1}{1}\)
\(\dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
\(\dfrac{\sin 3x}{\cos 3x} \cdot \dfrac{\cos 2x}{6x}\)
\(\dfrac{\sin 3x}{3x} \cdot \dfrac{3}{6} \cdot \dfrac{\cos 2x}{\cos 3x}\)
\(1 \cdot \dfrac{1}{2} \cdot 1\)
\(\dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
\(\dfrac{\sin 4x}{4x} \cdot \dfrac{4}{2} \cdot \dfrac{\cos 3x}{\cos 4x}\)
\(1 \cdot 2 \cdot 1\)
\(2\)
Answer: \(2\)
\(2 \cdot \dfrac{\sin 2x}{2x} \cdot \cos 3x\)
\(2 \cdot 1 \cdot 1\)
\(2\)
Answer: \(2\)
\(\dfrac{1 - \cos 4x}{x^2} = 16 \cdot \dfrac{1 - \cos 4x}{(4x)^2}\)
\(\lim_{\theta \to 0} \dfrac{1 - \cos \theta}{\theta^2} = \dfrac{1}{2}\)
\(16 \cdot \dfrac{1}{2}\)
\(8\)
Answer: \(8\)
\(\dfrac{\sin 5x}{5x} \cdot 5x \cdot \dfrac{2x}{\tan 2x} \cdot \dfrac{1}{2x}\)
\(\dfrac{\sin 5x}{5x} \cdot \dfrac{2x}{\tan 2x} \cdot \dfrac{5}{2}\)
\(1 \cdot 1 \cdot \dfrac{5}{2}\)
\(\dfrac{5}{2}\)
Answer: \(\dfrac{5}{2}\)
Need help with trig limits?
Book a Session with Amy