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:
\[{\lim\limits_{x \to 0} \frac{{\tan x}}{x} = 1,\;\;\;}\kern-0.3pt
{\lim\limits_{x \to 0} \frac{{\arcsin x}}{x} = 1,\;\;\;}\kern-0.3pt
{\lim\limits_{x \to 0} \frac{{\arctan x}}{x} = 1.}\] Further we assume that angles are measured in radians.

Solved Problems

Click on problem description to see solution.

Example 1.

Find the limit \(\lim\limits_{x \to 0} {\large\frac{{4x}}{{\sin 3x}}\normalsize}\).

Solution.

\[L
= {\lim\limits_{x \to 0} \frac{{4x}}{{\sin 3x}} }
= {\lim\limits_{x \to 0} \frac{{3 \cdot 4x}}{{3\sin 3x}} }
= {\frac{4}{3}\lim\limits_{x \to 0} \frac{{3x}}{{\sin 3x}} }
= {\frac{4}{3}\lim\limits_{x \to 0} \frac{1}{{\large\frac{{\sin 3x}}{{3x}}\normalsize}} }
= {\frac{4}{3}\frac{{\lim\limits_{x \to 0} 1}}{{\lim\limits_{x \to 0} \large\frac{{\sin 3x}}{{3x}}\normalsize}}.}
\] Since \(3x \to 0\) as \(x \to 0,\) we can write:
\[L
= \frac{4}{3}\frac{{\lim\limits_{x \to 0} 1}}{{\lim\limits_{x \to 0} \large\frac{{\sin 3x}}{{3x}}\normalsize}}
= \frac{4}{{3\lim\limits_{3x \to 0} \large\frac{{\sin 3x}}{{3x}}\normalsize}}
= \frac{4}{{3 \cdot 1}} = \frac{4}{3}.
\]

Example 2.

Calculate the limit \(\lim\limits_{x \to 0} {\large\frac{{\cos {3x} – \cos x}}{{{x^2}}}\normalsize}\).

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 \(\lim\limits_{x \to 0} {\large\frac{{\sin5x – \sin 3x}}{{\sin x}}\normalsize}\).

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
\[{\lim\limits_{x \to 0} \frac{{\sin5x – \sin 3x}}{{\sin x}} }
= {\lim\limits_{x \to 0} \frac{{2\sin \large\frac{{5x – 3x}}{2}\normalsize\cos \large\frac{{5x + 3x}}{2}\normalsize}}{{\sin x}} }
= {\lim\limits_{x \to 0} \frac{{2\sin x\cos 4x}}{{\sin x}} }
= {\lim\limits_{x \to 0} \left( {2\cos 4x} \right).}
\] As \(\cos{4x}\) is a continuous function at \(x = 0\), then
\[{\lim\limits_{x \to 0} \left( {2\cos 4x} \right) }
= {2\lim\limits_{x \to 0} \cos 4x }
= {2 \cdot \cos \left( {4 \cdot 0} \right) = 2 \cdot 1 = 2.}
\]