Question

How do you find the limit of `(x−pi/4)tan(2x)` as x approaches pi/4?

Answer 1

Please see the explanation section, below.

Explanation:

`lim_(xrarrpi/4) (x−pi/4)tan(2x)`

I'd like to use the fundamental trigonometric limts, so I'll change the variable.

Let `u = x-pi/4`,

so that `x = u+pi/4` and `xrarrpi/4` is equivalent to `urarr0`.

Nww we have been asked to find

`lim_(urarr0)utan2(u+pi/4) = lim_(urarr0)utan(2u+pi/2)`

Rewriting `tan` as `sin/cos`, we get

`= lim_(urarr0)usin(2u+pi/2)/cos(2u+pi/2)`

`= lim_(urarr0)u(sin(2u)cos(pi/2)+cos(2u)sin(pi/2))/(cos(2u)cos(pi/2)-sin(2u)sin(pi/2))`

`= lim_(urarr0)ucos(2u)/(-sin(2u))`

`= lim_(urarr0)(-1/2(2u)/sin(2u)cos(2u))`

`= -1/2(lim_(urarr0)(2u)/sin(2u))(lim_(urarr0)cos(2u))`

`= (-1/2)(1)(1) = -1/2`

`lim_(xrarrpi/4) (x−pi/4)tan(2x) = -1/2`

Answer 2

We have that

#lim_(xrarrpi/4) (x−pi/4)tan(2x)= lim_(xrarrpi/4) (x-pi/4)*sin(2x)/(cos2x)#

Hence the limit is the form `0/0` we apply L'Hopital Law hence

#lim_(xrarrpi/4) (d[(x-pi/4)*sin2x]/dx)/(d(cos2x)/dx)= lim_(xrarrpi/4) [sin2x+2*(x-pi/4)cos2x]/(-2sin2x)=-1/2#