Ucale

# Evaluation Of Trigonometric Limits (Form-1)

Limits of functions are evaluated using many different techniques such as recognizing a pattern, simple substitution, or using algebraic simplifications. Some of these techniques are illustrated in the following examples. $\displaystyle (i)\lim _{ x\longrightarrow 0 }{ \sin { \theta } } =0$

Let ABC be a right angled triangle such that $\displaystyle \angle C=\frac { \pi }{ 2 } \quad and\quad \angle ABC=\theta \\ Then\quad \sin { \theta } =\frac { CA }{ BA }$

Now , if we keep BC fixed and go on decreasing angle  θ then we find that  A goes on coming nearer and nearer to C  $\displaystyle \therefore A\longrightarrow C\quad as\quad \theta \longrightarrow 0$

This means that $\displaystyle CA\longrightarrow 0\quad as\quad \theta \longrightarrow 0\\ \Rightarrow \frac { CA }{ BA } \rightarrow 0\quad \quad as\quad \theta \longrightarrow 0\\ \Rightarrow \sin { \theta } \rightarrow 0\quad \quad as\quad \theta \longrightarrow 0\\ Thus,\lim _{ \theta \longrightarrow 0 }{ \sin { \theta } }=0$

Now, $\displaystyle \left( ii \right) \lim _{ \theta \longrightarrow 0 }{ \sin { \theta } } =1$ $\displaystyle BA\longrightarrow BC\quad as\quad \theta \longrightarrow 0\\ \Rightarrow \frac { BC }{ BA } \rightarrow 0\quad \quad as\quad \theta \longrightarrow 0\\ \Rightarrow \cos { \theta } \rightarrow 1\quad \quad as\quad \theta \longrightarrow 0\\ Thus,\lim _{ \theta \longrightarrow 0 }{ \cos { \theta } } =1$ $\displaystyle \left( iii \right) \lim _{ x\longrightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } =1$

Consider a circle of radius r. Let O be the centre of the circle such that $\displaystyle \angle AOB=\theta$

where $\displaystyle \theta$ is measured in radians and it is very small . Suppose the tangent at A meets OB produced at P . From the below figure , we have  $\displaystyle Area\quad of\quad \Delta OAB\textless Area\quad of\quad sector\quad OAB\textless Area\quad of\quad \Delta OAP\\ \Rightarrow \frac { 1 }{ 2 } OA.OB\sin { \theta } \textless \frac { 1 }{ 2 } { \left( OA \right) }^{ 2 }\theta \\\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \textless \frac { 1 }{ 2 } OA.AP\\ \Rightarrow \frac { 1 }{ 2 } { r }^{ 2 }\sin { \theta } \textless \frac { 1 }{ 2 } { r }^{ 2 }\theta \textless \frac { 1 }{ 2 } { r }^{ 2 }\tan { \theta } \qquad \left[ In\quad \Delta OAP,AP=OA\tan { \theta } \right] \\ \Rightarrow \sin { \theta } \textless \theta \textless \tan { \theta } \\ \Rightarrow 1\textless \frac { \theta }{ \sin { \theta } } \textless \frac { 1 }{ \cos { \theta } } \qquad \left[ \because \theta \quad is\quad small\quad \therefore \sin { \theta } \textgreater 0 \right] \\ \Rightarrow 1 \textgreater \frac { \sin { \theta } }{ \theta } \textgreater \cos { \theta } \\ \Rightarrow 1\ge \lim _{ \theta \longrightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } \ge \lim _{ \theta \longrightarrow 0 }{ \cos { \theta } } \quad or\quad \lim _{ \theta \longrightarrow 0 }{ \cos { \theta } } \le \lim _{ \theta \longrightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } \le 1\\ \Rightarrow 1\le \lim _{ \theta \longrightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } \le 1\\ \Rightarrow \lim _{ \theta \longrightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } =1$ $\displaystyle \left( iv \right) \lim _{ \theta \longrightarrow 0 }{ \frac { \tan { \theta } }{ \theta } } \\ =\lim _{ \theta \longrightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } .\frac { 1 }{ \cos { \theta } } \\ =\lim _{ \theta \longrightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } .\lim _{ \theta \longrightarrow 0 }{ \frac { 1 }{ \cos { \theta } } } \\ =\left( 1 \right) .\left( 1 \right) \\ =1$ $\displaystyle \left( v \right) \lim _{ \theta \longrightarrow a }{ \frac { \sin { \left( \theta -a \right) } }{ \theta -a } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { \sin { \left( a+h-a \right) } }{ a+h-a } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { \sin { h } }{ h } } \\ =1$ $\displaystyle \left( vi \right) \lim _{ \theta \longrightarrow a }{ \frac { \tan { \left( \theta -a \right) } }{ \theta -a } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { \tan { \left( a+h-a \right) } }{ a+h-a } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { \tan { h } }{ h } } \\ =1$

### Example $\displaystyle \left( i \right) \lim _{ x\longrightarrow 0 }{ \frac { \sin { 3x } }{ x } } \\ =\lim _{ x\longrightarrow 0 }{ \left( 3.\frac { \sin { 3x } }{ 3x } \right) } \\ =3\lim _{ x\longrightarrow 0 }{ \left( \frac { \sin { 3x } }{ 3x } \right) } \\ =3\left( 1 \right) \left[ \because \lim _{ x\longrightarrow 0 }{ \left( \frac { \sin { \theta } }{ \theta } \right) } =1 \right] \\ =3$

April 18, 2019
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