Question

lim_(x->pi/2) (x-(pi/2))tanx
ok so, i applied L'Hopital's Rule and got
lim_(x->pi/2) tanx+(x-pi/2)sec^2(x)
Now, plugin pi/2 and I got 1.
The answer is -1
WolfRam said it was -1 too, but couldn't show me the steps. Thanks in advance :)

Answer 1

lim as x --> pi/2 of { (x-pi/2) / (cosx/sinx) }

Answer 2

that doesn't help...

Answer 3

When applying L'Hopitals Rule I believe you need to form a fraction
rewrite expression
\[(x-\pi/2)\tan x = \frac{x- \pi/2}{\cot x}\]
\[\frac{d}{dx} \cot x = -\frac{1}{\sin^{2} x}\]
\[\rightarrow \lim_{x \rightarrow \pi/2} \frac{1}{-\frac{1}{\sin^{2} x}} = -\sin^{2} x = -1\]

Answer 4

how did you get \[(x-\pi/2)/\cot(x)\]?

Answer 5

from the fact that
\[\frac{1}{\cot x} = \tan x\]

Answer 6

\[\cot x = \frac{1}{\tan x}\]

Answer 7

ahh, ok. Thanks

Answer 8

your welcome