Trigonometric Limits
The basic trigonometric limit is
\[\lim\limits_{x \to 0} \frac{{\sin x}}{x} = 1.\]
Using this limit, one can get the series of other trigonometric limits:
\[\lim\limits_{x \to 0} \frac{{\tan x}}{x} = 1,\;\;\; \lim\limits_{x \to 0} \frac{{\arcsin x}}{x} = 1,\;\;\; \lim\limits_{x \to 0} \frac{{\arctan x}}{x} = 1.\]
Further we assume that angles are measured in radians.
Solved Problems
Example 1.
Find the limit \[\lim\limits_{x \to 0} {\frac{{4x}}{{\sin 3x}}}.\]
Solution.
\[L = \lim\limits_{x \to 0} \frac{{4x}}{{\sin 3x}} = \lim\limits_{x \to 0} \frac{{3 \cdot 4x}}{{3\sin 3x}} = \frac{4}{3}\lim\limits_{x \to 0} \frac{{3x}}{{\sin 3x}} = \frac{4}{3}\lim\limits_{x \to 0} \frac{1}{{\frac{{\sin 3x}}{{3x}}}} = \frac{4}{3}\frac{{\lim\limits_{x \to 0} 1}}{{\lim\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}}}.\]
Since \(3x \to 0\) as \(x \to 0,\) we can write:
\[L = \frac{4}{3}\frac{{\lim\limits_{x \to 0} 1}}{{\lim\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}}} = \frac{4}{{3\lim\limits_{3x \to 0} \frac{{\sin 3x}}{{3x}}}} = \frac{4}{{3 \cdot 1}} = \frac{4}{3}.\]
Example 2.
Calculate the limit \[\lim\limits_{x \to 0} {\frac{{\cos {3x} - \cos x}}{{{x^2}}}}.\]
Solution.
We factor the numerator:
\[\cos{3x} - \cos x = - 2\sin \frac{{3x - x}}{2}\sin \frac{{3x + x}}{2} = - 2\sin x\sin 2x. \]
This yields
\[\lim\limits_{x \to 0} \frac{{\cos 3x - \cos x}}{{{x^2}}} = \lim\limits_{x \to 0} \frac{{\left( { - 2\sin x\sin 2x} \right)}}{{{x^2}}} = - 2\lim\limits_{x \to 0} \frac{{\sin x}}{x} \cdot \lim\limits_{x \to 0} \frac{{\sin 2x}}{x} = - 2 \cdot 1 \cdot \lim\limits_{2x \to 0} \frac{{2\sin 2x}}{{2x}} = - 2 \cdot 2\lim\limits_{2x \to 0} \frac{{\sin 2x}}{{2x}} = - 4.\]
Example 3.
Find the limit \[\lim\limits_{x \to 0} {\frac{{\sin5x - \sin 3x}}{{\sin x}}}.\]
Solution.
We use the following trigonometric identity:
\[\sin x - \sin y = 2\sin \frac{{x - y}}{2}\cos \frac{{x + y}}{2}.\]
Then we obtain
\[\lim\limits_{x \to 0} \frac{{\sin5x - \sin 3x}}{{\sin x}} = \lim\limits_{x \to 0} \frac{{2\sin \frac{{5x - 3x}}{2} \cos \frac{{5x + 3x}}{2}}}{{\sin x}} = \lim\limits_{x \to 0} \frac{{2\sin x\cos 4x}}{{\sin x}} = \lim\limits_{x \to 0} \left( {2\cos 4x} \right).\]
As \(\cos{4x}\) is a continuous function at \(x = 0,\) then
\[\lim\limits_{x \to 0} \left( {2\cos 4x} \right) = 2\lim\limits_{x \to 0} \cos 4x = 2 \cdot \cos \left( {4 \cdot 0} \right) = 2 \cdot 1 = 2.\]
Example 4.
Calculate the limit \[\lim\limits_{x \to 0} {\frac{{\cos \left( {x + a} \right) - \cos \left( {x - a} \right)}}{x}}.\]
Solution.
Using the trig identity
\[\cos \alpha - \cos \beta = - 2\sin \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2},\]
we convert the limit in the following way:
\[L = \lim\limits_{x \to 0} \frac{{\cos \left( {x + a} \right) - \cos \left( {x - a} \right)}}{x} = - 2\lim\limits_{x \to 0} \frac{{\sin x\sin a}}{x}.\]
Here \(\sin a\) is a constant that does not depend on \(x.\) Therefore,
\[L = - 2\sin a\,\underbrace {\lim\limits_{x \to 0} \frac{{\sin x}}{x}}_1 = - 2\sin a.\]
See more problems on Page 2.
