Question

Please solve giving step-by-step detail:
lim x-> pi/2 (tanx(^cosx))

Answer 1

Let A=lim x->pi/2 (tanx)^cosx. Put both sides inside ln function:
\[\ln A=\ln[\lim_{x \rightarrow \frac{ \pi }{ 2 }}\tan x ^{\cos x}]\]
Limit and ln function can be replaced:
\[\ln [\lim_{x \rightarrow \frac{ \pi }{ 2 }} \tan x ^{\cos x}]=\lim_{x \rightarrow \frac{ \pi }{ 2 }}[\ln (\tan x)^{\cos x}]\]
Use now property of ln function:
\[\ln Y ^{Z}=Z*\ln Y\]
\[\lim_{x \rightarrow \frac{ \pi }{ 2 }}[(\cos x)*\ln \tan x]\]. You can write this as:
\[\lim_{x \rightarrow \frac{ \pi }{ 2 }}[\frac{ \ln \tan x }{ \cos x}]\]
These form is inf/inf and you can aply l'hopital rule:
\[=\lim_{x \rightarrow \frac{ \pi }{ 2 }}\frac{ \frac{ 1 }{ \tan x* \cos ^{2} x} }{ \frac{ \sin x }{ \cos ^{2}x }}\]
Simplyfing these:\[\lim_{x \rightarrow \frac{ \pi }{ 2 }}\frac{ \cos x }{ \sin ^{2} x}=0\]
And now recall that ln(A)=0, where A is solution to given limit. Function ln(Y) is equal to
0 only if its argument Y is equal to 1, so we get that A=1 and solution to given limit
is 1.

Answer 2

How did u get (ln tanx)/cos x ??

Answer 3

I made mistake in typing, correct is:\[\frac{ \ln \tan x }{ \frac{ 1 }{cosx} }\]