Question

Anyone solve this
$lim_{x \rightarrow \frac{\pi}{2}} \frac{cot(x) -cos(x) }{{\pi - 2x)}^{3}}$
equals ?​

Answer 1

Answer:

$\boxed{\bold{\mathsf{= \frac{1}{16}}}}$

Step-by-step explanation:

Given:-

$lim_{x \rightarrow \frac{\pi}{2}} \frac{cot(x) -cos(x) }{{\pi - 2x)}^{3}}$

$\rule{200}{1}$

Solve as below :-

Divide by 2 inside

and taking 1/8 as common

$\mathsf{lim_{x \rightarrow \frac{\pi}{2}} \frac{1}{8} \times \frac{cosx(1-sinx)}{sinx \times({\frac{\pi}{2}-x)}^{3}}}$

now , reduce the limit

$\mathsf{lim_{h \rightarrow 0} \frac{1}{8} \times \frac{cos(\frac{\pi}{2} - h) \times \left[1-sin(\frac{\pi}{2} -h)\right]}{sin(\frac{\pi}{2}- h)\times {(\frac{\pi}{2}-\frac{\pi}{2}+h)}^{3}}}$

$\mathsf{ \frac{1}{8}\times lim_{ h \rightarrow 0} \frac{sinh(1-cosh)}{cosh \times {h}^{3}}}$

$\mathsf{ \frac{1}{8}\times lim_{h \rightarrow 0} \frac{sinh(2{sin}^{2}\frac{h}{2})}{cosh\times {h}^{3}}}$

Taking out 1/2 wholey as common

$\mathsf{ \frac{1}{4}\times lim_{h \rightarrow 0}\frac{sinh\times{sin}^{2}\frac{h}{2})}{{h}^{3}cosh}}$

$\frac{1}{4} \times lim_{h \rightarrow 0} \frac{(sinh)}{h}\times (\frac{sin(\frac{h}{2})}{{\frac{h}{2}}}^{2} \times \frac{1}{cosh}\times \frac{1}{4}$

$\rule{200}{1}$

Solving out the limit

$\bold{\mathsf{\frac{1}{4}\times\frac{1}{4}}}$

$\bold{\mathsf{= \frac{1}{16}}}$

$\rule{200}{1}$

Important:

$\boxed{\bold{\mathsf{\frac{sinx}{x}= 1}}}$

$\boxed{\bold{\mathsf{(sin\frac{\pi}{2} - a)=cos(a)}}}$

Answer 2

Answer:

Answer:

\boxed{\bold{\mathsf{= \frac{1}{16}}}}

=

16

1

Step-by-step explanation:

Given:-

lim_{x \rightarrow \frac{\pi}{2}} \frac{cot(x) -cos(x) }{{\pi - 2x)}^{3}} lim

x→

2

π

π−2x)

3

cot(x)−cos(x)

\rule{200}{1}

Solve as below :-

Divide by 2 inside

and taking 1/8 as common

\mathsf{lim_{x \rightarrow \frac{\pi}{2}} \frac{1}{8} \times \frac{cosx(1-sinx)}{sinx \times({\frac{\pi}{2}-x)}^{3}}}lim

x→

2

π

8

1

×

sinx×(

2

π

−x)

3

cosx(1−sinx)

now , reduce the limit

\mathsf{lim_{h \rightarrow 0} \frac{1}{8} \times \frac{cos(\frac{\pi}{2} - h) \times \left[1-sin(\frac{\pi}{2} -h)\right]}{sin(\frac{\pi}{2}- h)\times {(\frac{\pi}{2}-\frac{\pi}{2}+h)}^{3}}}lim

h→0

8

1

×

sin(

2

π

−h)×(

2

π

−

2

π

+h)

3

cos(

2

π

−h)×[1−sin(

2

π

−h)]

\mathsf{ \frac{1}{8}\times lim_{ h \rightarrow 0} \frac{sinh(1-cosh)}{cosh \times {h}^{3}}}

8

1

×lim

h→0

cosh×h

3

sinh(1−cosh)

\mathsf{ \frac{1}{8}\times lim_{h \rightarrow 0} \frac{sinh(2{sin}^{2}\frac{h}{2})}{cosh\times {h}^{3}}}

8

1

×lim

h→0

cosh×h

3

sinh(2sin

2

2

h

)

Taking out 1/2 wholey as common

\mathsf{ \frac{1}{4}\times lim_{h \rightarrow 0}\frac{sinh\times{sin}^{2}\frac{h}{2})}{{h}^{3}cosh}}

4

1

×lim

h→0

h

3

cosh

sinh×sin

2

2

h

)

\frac{1}{4} \times lim_{h \rightarrow 0} \frac{(sinh)}{h}\times (\frac{sin(\frac{h}{2})}{{\frac{h}{2}}}^{2} \times \frac{1}{cosh}\times \frac{1}{4}

4

1

×lim

h→0

h

(sinh)

×(

2

h

sin(

2

h

)

2

×

cosh

1

×

4

1

\rule{200}{1}

Solving out the limit

\bold{\mathsf{\frac{1}{4}\times\frac{1}{4}}}

4

1

×

4

1

\bold{\mathsf{= \frac{1}{16}}}=

16

1

\rule{200}{1}

Important:

\boxed{\bold{\mathsf{\frac{sinx}{x}= 1}}}

x

sinx

=1

\boxed{\bold{\mathsf{(sin\frac{\pi}{2} - a)=cos(a)}}}

(sin

2

π

−a)=cos(a)

i hope its help u❤