Evaluate the following limit: lim x--> pi/2 (x - pi/2) * tanx Show work please!?

Hello,

lim {[x - (π/2)] tanx} =

x→π/2

the limit is of the form 0 ∙ ∞

let's rewrite tanx in terms of sinx, cosx:

lim {[x - (π/2)] (sinx /cosx)} =

x→π/2

let's rearrange this as:

lim {sinx {[x - (π/2)] /cosx} } =

x→π/2

lim {sinx /{cosx /[x - (π/2)]} } =

x→π/2

let's apply the co-function identity cosx = sin[(π/2) - x]:

lim {sinx /{{sin[(π/2) - x]} /[x - (π/2)]} } =

x→π/2

let's change signs as:

lim {sinx /{{sin[(π/2) - x]} /{- [- x + (π/2)]}} } =

x→π/2

lim {sinx /{- {{sin[(π/2) - x]} /[(π/2) - x]}} } =

x→π/2

let's note that {sin[(π/2) - x]} /[(π/2) - x] is a notable limit of the form lim [(sin t) /t] = 1, in

..................... ................. .......................... ...................... .................t→0

that, if we let (π/2) - x = t, as x approaches π/2, t tends to zero; then, letting x approach π/2 we obtain:

lim {sin(→π/2) /[- (→1)]} =

x→π/2

lim [(→1) /(→ - 1)] =

x→π/2

- 1

I hope it helps

x = pi/2 - u

u = pi/2 - x

lim u--> 0 (-u) * tan(pi/2 - u)

lim u--> 0 (-u) * cot(u)

lim u--> 0 (-u) * cos(u) / sin u

lim u--> 0 (-u/ sin u) * cos(u)

= -1 * 1 = -1