Question

\[\lim_{x \rightarrow \frac{\pi}{2}}(sinx)^{tanx}\]

Let f(x) = (sinx)^(tanx)
lnf(x) = (tanx) ln (sinx)
So, \[\lim_{x \rightarrow \frac{\pi}{2}} lnf(x) = \lim_{x \rightarrow \frac{\pi}{2}}(tanx) ln (sinx)\]
Then, I don't know how to continue..

Answer 1

\[\tan x = \frac{\sin x}{\cos x}\]then apply L'hopital rule

Answer 2

but \[\lim_{x \rightarrow \frac{\pi}{2}}\frac{sinx}{cosx} \] would be like 1 / 0, can we use L'Hopital rule in this case?

Answer 3

I was told that I can only use L'Hopital rule when it's in the form ∞/∞, or 0/0

Answer 4

wait, it is not 1/0\[\lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin x \ln \left( \sin x \right)}{\cos x} = \frac{0}{0}\]

Answer 5

Oh!! My bad!! Thanks!!