Trigonometric Limits
Solved Problems
Example 5.
Calculate the limit \[\lim\limits_{x \to 0} {\frac{{\sin ax}}{{\sin bx}}}.\]
Solution.
\[L = \lim\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \lim\limits_{x \to 0} \left( {\frac{{\sin ax}}{{\sin bx}} \cdot \frac{a}{b} \cdot \frac{{bx}}{{ax}}} \right) = \lim\limits_{x \to 0} \left( {\frac{{\sin ax}}{{ax}} \cdot \frac{{bx}}{{\sin bx}} \cdot \frac{a}{b}} \right) = \frac{a}{b}\frac{{\lim\limits_{x \to 0} \frac{{\sin ax}}{{ax}}}}{{\lim\limits_{x \to 0} \frac{{\sin bx}}{{bx}}}}.\]
Obviously, \(ax \to 0\) and \(bx \to 0\) as \(x \to 0.\) Then
\[L = \frac{a}{b}\frac{{\lim\limits_{x \to 0} \frac{{\sin ax}}{{ax}}}}{{\lim\limits_{x \to 0} \frac{{\sin bx}}{{bx}}}} = \frac{a}{b} \cdot \frac{1}{1} = \frac{a}{b}.\]
Example 6.
Find the limit \[\lim\limits_{x \to b} {\frac{{\sin x - \sin b}}{{x - b}}}.\]
Solution.
Using the identity
\[\sin x - \sin b = 2\sin \frac{{x - b}}{2}\cos\frac{{x + b}}{2},\]
convert the limit:
\[L = \lim\limits_{x \to b} \frac{{\sin x - \sin b}}{{x - b}} = \lim\limits_{x \to b} \frac{{2\sin \frac{{x - b}}{2} \cos \frac{{x + b}}{2}}}{{x - b}} = \lim\limits_{x \to b} \frac{{\sin \frac{{x - b}}{2}}}{{\frac{{x - b}}{2}}} \cdot \lim\limits_{x \to b} \cos \frac{{x + b}}{2} = \lim\limits_{x \to b} \frac{{\sin \frac{{x - b}}{2}}}{{\frac{{x - b}}{2}}} \cdot \cos b.\]
Change the variable: \({\frac{{x - b}}{2}} = t.\) When \(x \to b,\) then \(2t \to 0\) or \(t \to 0.\) Hence,
\[L = \lim\limits_{t \to 0} \frac{{\sin t}}{t} \cdot \cos b = 1 \cdot \cos b = \cos b.\]
Example 7.
Find the limit \[\lim\limits_{x \to 0} {\frac{{\tan x - \sin x}}{{{x^3}}}}.\]
Solution.
We apply the following transformations:
\[L = \lim\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}} = \lim\limits_{x \to 0} \frac{{\frac{{\sin x}}{{\cos x}} - \sin x}}{{{x^3}}} = \lim\limits_{x \to 0} \frac{{\sin x\left( {\frac{1}{{\cos x}} - 1} \right)}}{{{x^3}}} = \lim\limits_{x \to 0} \frac{{\sin x\left( {1 - \cos x} \right)}}{{{x^3}\cos x}} = \lim\limits_{x \to 0} \left[ {\frac{{\sin x}}{x} \cdot \frac{{1 - \cos x}}{{{x^2}\cos x}}} \right].\]
As \(1 - \cos x = 2{\sin ^2}{\frac{x}{2}},\) we have
\[L = \lim\limits_{x \to 0} \left[ {\frac{{\sin x}}{x} \cdot \frac{{1 - \cos x}}{{{x^2}\cos x}}} \right] = \lim\limits_{x \to 0} \left[ {\frac{{\sin x}}{x} \cdot \frac{{2{{\sin }^2} \frac{x}{2}}}{{{x^2}\cos x}}} \right] = \frac{{\lim\limits_{x \to 0} \frac{{\sin x}}{x} \cdot 2\lim\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}}}}{{\lim\limits_{x \to 0} \cos x}}.\]
Here
\[\lim\limits_{x \to 0} \frac{{\sin x}}{x} = 1\;\;\text{and}\;\;\; \lim\limits_{x \to 0} \cos x = 1.\]
Hence,
\[L = 2\lim\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}} = 2\lim\limits_{x \to 0} \left( {\frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}} \cdot \frac{4}{4}} \right) = 2\lim\limits_{x \to 0} \left( {\frac{{{{\sin }^2}\frac{x}{2}}}{{\frac{{{x^2}}}{4}}} \cdot \frac{1}{4}} \right) = \frac{1}{2}\lim\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{{\left( {\frac{x}{2}} \right)}^2}}}.\]
Here \({\frac{x}{2}} \to 0\) as \(x \to 0.\) Therefore
\[L = \frac{1}{2}\lim\limits_{\frac{x}{2} \to 0} {\left( {\frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)^2} = \frac{1}{2} \cdot {1^2} = \frac{1}{2}.\]
Example 8.
Find the limit \[\lim\limits_{x \to {\frac{1}{2}}} {\frac{{1 - 4{x^2}}}{{\arcsin \left( {1 - 2x} \right)}}}.\]
Solution.
We make the substitution: \(x - {\frac{1}{2}} = y.\) When \(x \to {\frac{1}{2}},\) then \(y \to 0.\) As a result, we have
\[L = \lim\limits_{x \to \frac{1}{2}} \frac{{1 - 4{x^2}}}{{\arcsin \left( {1 - 2x} \right)}} = \lim\limits_{y \to 0} \frac{{1 - 4{{\left( {y + \frac{1}{2}} \right)}^2}}}{{\arcsin \left( {1 - 2\left( {y + \frac{1}{2}} \right)} \right)}} = \lim\limits_{y \to 0} \frac{{1 - 4\left( {{y^2} + y + \frac{1}{4}} \right)}}{{\arcsin \left( {1 - 2y - 1} \right)}} = \lim\limits_{y \to 0} \frac{{1 - 4{y^2} - 4y - 1}}{{\arcsin \left( { - 2y} \right)}} = \lim\limits_{y \to 0} \frac{{ - 4{y^2} - 4y}}{{\arcsin \left( { - 2y} \right)}} = \lim\limits_{y \to 0} \frac{{ - 4{y^2} - 4y}}{{\frac{{\arcsin \left( { - 2y} \right)}}{{ - 2y}}\ \cdot \left( { - 2y} \right)}} = \frac{{\lim\limits_{y \to 0} \frac{{ - 4{y^2} - 4y}}{{ - 2y}}}}{{\lim\limits_{y \to 0} \frac{{\arcsin \left( { - 2y} \right)}}{{ - 2y}}}} = \frac{{\lim\limits_{y \to 0} \left( {2y + 2} \right)}}{{\lim\limits_{ - 2y \to 0} \frac{{\arcsin \left( { - 2y} \right)}}{{\left( { - 2y} \right)}}}} = \frac{{\lim\limits_{y \to 0} \left( {2y + 2} \right)}}{1} = \lim\limits_{y \to 0} \left( {2y + 2} \right) = 2. \]
Example 9.
Find the limit \[\lim\limits_{x \to 0 + 0} {\frac{{\sqrt {1 - \cos x} }}{x}}.\]
Solution.
We use the trigonometric formula:
\[1 - \cos x = 2{\sin ^2}\frac{x}{2}.\]
Then the limit can be written in the form
\[L = \lim\limits_{x \to 0 + 0} \frac{{\sqrt {1 - \cos x} }}{x} = \lim\limits_{x \to 0 + 0} \frac{{\sqrt {2{{\sin }^2}\frac{x}{2}} }}{x} = \sqrt 2 \lim\limits_{x \to 0 + 0} \frac{{\sqrt {{{\sin }^2}\frac{x}{2}} }}{x} = \sqrt 2 \lim\limits_{x \to 0 + 0} \sqrt {\frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}}} = \sqrt 2 \lim\limits_{x \to 0 + 0} \sqrt {\frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}} \cdot \frac{4}{4}} = \sqrt 2 \lim\limits_{x \to 0 + 0} \left( {\sqrt {\frac{{{{\sin }^2}\frac{x}{2}}}{{\frac{{{x^2}}}{4}}}} \cdot \frac{1}{{\sqrt 4 }}} \right) = \frac{{\sqrt 2 }}{2}\lim\limits_{x \to 0 + 0} \sqrt {\frac{{{{\sin }^2}\frac{x}{2}}}{{{{\left( {\frac{x}{2}} \right)}^2}}}} = \frac{1}{{\sqrt 2 }}\sqrt {\lim\limits_{x \to 0 + 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{{\left( {\frac{x}{2}} \right)}^2}}}} = \frac{1}{{\sqrt 2 }}\sqrt {{{\left[ {\lim\limits_{\frac{x}{2} \to 0 + 0} \left( {\frac{{\sin \frac{x}{2}}}{{\frac{x}{2}}}} \right)} \right]}^2}} = \frac{1}{{\sqrt 2 }} \cdot \sqrt {{1^2}} = \frac{1}{{\sqrt 2 }}.\]