Calculus

Limits and Continuity of Functions

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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

Click or tap a problem to see the solution.

Example 1

Find the limit \[\lim\limits_{x \to 0} {\frac{{4x}}{{\sin 3x}}}.\]

Example 2

Calculate the limit \[\lim\limits_{x \to 0} {\frac{{\cos {3x} - \cos x}}{{{x^2}}}}.\]

Example 3

Find the limit \[\lim\limits_{x \to 0} {\frac{{\sin5x - \sin 3x}}{{\sin x}}}.\]

Example 4

Calculate the limit \[\lim\limits_{x \to 0} {\frac{{\cos \left( {x + a} \right) - \cos \left( {x - a} \right)}}{x}}.\]

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.

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