Question

How do you find the limit of `(x-pi/2)tanx` as `x->pi/2`?

Answer 1

Rewrite so we can use `lim_(thetararr0)theta/sintheta = 1`

Explanation:

`cos(pi/2-x) = sinx` (by co-function identity or difference formula for cosine)

Also `cos(-theta) = cos theta` (cosine is an even function)

So `cos(x-pi/2) = cos(pi/2-x) = sinx` and

sinx = cos(x-pi/2)

`sin(pi/2-x) = cosx` (by co-function identity or difference formula for sine)

Also `sin(-theta) = -sin theta` (sine is an odd function)

So `sin(x-pi/2) = -sin(pi/2-x) = -cosx`and

cosx = -sin(x-pi/2)

Now we can rewrite

`lim_(xrarrpi/2)(pi/2-x)tanx = lim_(xrarrpi/2)(pi/2-x)sinx/cosx`

`= lim_(xrarrpi/2)(pi/2-x)cos(pi/2-x)/(-sin(pi/2-x))`

`= lim_(xrarrpi/2)-(pi/2-x)/(sin(pi/2-x)) * cos(pi/2-x)`

Notice that `lim_(xrarrpi/2)(pi/2-x)/(sin(pi/2-x))` is a version of `lim_(thetararr0)theta/sintheta`, so we get

`= -(1)(1) = -1`