Question

(In the form of 0*infinity) Find the limit as x approaches pi/2 from the negative of (pi/2-x)secx

Answer 1

\[\lim_{x \rightarrow \frac{ \pi }{ 2 }-}\left( \frac{ \pi }{2 }-x \right)\sec x\]
\[put~x=\frac{ \pi }{2 }-h,h>0,h \rightarrow0~asx \rightarrow \frac{ \pi }{ 2 }-\]
\[\lim_{h \rightarrow 0}\left\{ \frac{ \pi }{2 }-\left( \frac{ \pi }{ 2 }-h \right) \right\}\sec \left( \frac{ \pi }{ 2 }-h \right)\]
\[\lim_{h \rightarrow 0}h \csc h=\lim_{h \rightarrow0}\frac{ h }{ \sin h }=?\]

Answer 2

Why does limh→0{π2−(π2−h)}sec(π2−h) turn into limh→0hcsch? Was this done with differentiation?

Answer 3

\[\sec \left( \frac{ \pi }{ 2 }-h \right)=\frac{ 1 }{\cos \left( \frac{ \pi }{ 2 }-h \right) }=\frac{ 1 }{\cos \frac{ \pi }{ 2 } \cos h+\sin \frac{ \pi }{2 }\sin h} \]
\[\cos \frac{ \pi }{2 }=0,0*\cos h=0,\sin \frac{ \pi }{ 2 }=1\]

Answer 4

\[=\frac{ 1 }{ \sin h }=\csc h\]

Answer 5

general rule to remember\[\frac{ \pi }{ 2 }-x~ is~add ~co~if~absent,~delete~co~if~present\]
for example sine in cosine,here co is added.
secant to cosecant,here again co is added.
cotangent to tangent,here co is deleted.
cosine to sine ,here co is deleted.

Answer 6

That means:
sine(pi/2-x) = cosine(x)
tangent(pi/2-x) = cotangent(x)
secant(pi/2-x) = cosecant(x)

Answer 7

And vice versa

Answer 8

correct