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

2 Answers
May 26, 2016

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#

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#