Limits · Definition
Special trigonometric limits
Before you start you need to know
Two limits with sine and cosine that help solve many other trigonometric limits.
Definition
Two basic limits in radians: \(\frac{\sin x}{x}\) tends to \(1\) and \(\frac{1-\cos x}{x}\) tends to \(0\) as \(x\) tends to \(0\).
A few limits are proved once (with the squeeze theorem and geometry) and reused constantly. The most famous is \(\lim_{x\to0}\frac{\sin x}{x}=1\), together with its relative \(\lim_{x\to0}\frac{1-\cos x}{x}=0\).
Their importance is that the derivatives of sine and cosine come from them. Without this limit there would be no formula for the derivative of \(\sin x\).
A common trick: for \(\lim_{x\to0}\frac{\sin(kx)}{x}\) multiply and divide by \(k\) to recognize \(\frac{\sin u}{u}\) with \(u=kx\).
Think about it
What is \(\lim_{x\to0}\frac{\sin 3x}{x}\)?
It is 3: rewrite it as \(3\cdot\frac{\sin 3x}{3x}\) and the factor tends to 1.
Exercises 11
Compute the following limits
- \(\displaystyle \lim_{x\to 0}\,\frac{\sin {3 x}}{x}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{\sin {5 x}}{2 x}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{\tan {4 x}}{x}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{1 - \cos {x}}{x^{2}}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{\sin {2 x}}{\sin {3 x}}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{\sin {7x}}{x}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{\sin {4x}}{8x}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{1-\cos {2x}}{x^{2}}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{x}{\sin {3x}}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{\tan {2x}}{\sin {5x}}\)
- \(\displaystyle \lim_{x\to 0}\,\frac{\sin {x^{2}}}{1-\cos {x}}\)
What comes next
References
- James Stewart, Calculus: Early Transcendentals, 8.ª edición, Cengage Learning, 2015.