Ucale

# Evaluation Of Trigonometric Limits (Form-2)

So far we were discussing trigonometric limits when $\displaystyle x\longrightarrow 0$ . Now we will discuss evaluation of trigonometric limits when x tends to non-zero real number. As we have already assumed that $\displaystyle \lim _{ x\longrightarrow a }{ f\left( x \right) }$ always exists, therefore $\displaystyle \lim _{ x\longrightarrow a }{ f\left( x \right) } =\lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) } \quad \left[ \because \lim _{ x\longrightarrow { a }^{ - } }{ f\left( x \right) } =\lim _{ x\longrightarrow { a }^{ + } }{ f\left( x \right) } \Leftrightarrow \lim _{ x\longrightarrow a }{ f\left( x \right) } \right] \\ =\lim _{ h\longrightarrow 0 }{ f(a+h) }$

Thus, if $\displaystyle x\longrightarrow a\left( a\neq 0 \right) ,then\quad we\quad replace\quad x\quad by\quad a+h\quad and\quad x\longrightarrow a\quad by\quad h\longrightarrow 0$

### Example $\displaystyle \lim _{ x\longrightarrow \frac { \pi }{ 2 } }{ \frac { 1+\cos { 2x } }{ { \left( \pi -2x \right) }^{ 2 } } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { 1+\cos { 2\left( \frac { \pi }{ 2 } +h \right) } }{ \left[ \pi -2\left( \frac { \pi }{ 2 } +h \right) \right] } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { 1+\cos { \left( \pi +2h \right) } }{ 4{ h }^{ 2 } } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { 1-\cos { \left( 2h \right) } }{ 4{ h }^{ 2 } } } \\ =\lim _{ h\longrightarrow 0 }{ \frac { 2{ sin }^{ 2 }h }{ 4{ h }^{ 2 } } } \\ =\frac { 2 }{ 4 } \lim _{ h\longrightarrow 0 }{ \frac { \sin { h } }{ h } } .\frac { \sin { h } }{ h } \\ =\frac { 1 }{ 2 } \left( \lim _{ h\longrightarrow 0 }{ \frac { \sin { h } }{ h } } \right) \left( \lim _{ h\longrightarrow 0 }{ \frac { \sin { h } }{ h } } \right) \\ =\frac { 1 }{ 2 }$

April 18, 2019
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