Anyone solve this
equals ?
Answers
Answer:
Step-by-step explanation:
Given:-
Solve as below :-
Divide by 2 inside
and taking 1/8 as common
now , reduce the limit
Taking out 1/2 wholey as common
Solving out the limit
Important:
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)