How do you find the limit of #tan(4alpha)/sin(2alpha)# as #alpha->0#?

#lim_(alpha to 0) (tan4alpha)/(sin 2alpha)#

This is in #0/0# indeterminate form so L'Hopital is a possibility.

We do it algebraically, using:

#sin 2 z = 2 sin z cos z#, and

#cos 2z = cos^2 z - sin^2 z# and variations of this and the Pythagorean ID.

We can say:

#= lim_(alpha to 0) (sin 4alpha)/(cos 4 alpha) 1/(sin 2alpha)#

#= lim_(alpha to 0) (2 sin 2alpha cos 2 alpha)/(2 cos^2 2alpha - 1) 1/(sin 2alpha)#

#= lim_(alpha to 0) (2 cos 2 alpha)/(2 cos^2 2alpha - 1) #

These are both continuous at #alpha = 0# so we can plug in #alpha = 0# to get:

#= lim_(alpha to 0) (2 * 1)/(2 *1 - 1) = 2 #

#tan(4alpha) = sin(4alpha)/cos(4alpha)#

Use the sine of a double angle formula to rewrite this

# = (2sin(2alpha)cos(2alpha))/cos(4alpha)#

So

#tan(4alpha)/sin(2alpha) = (2sin(2alpha)cos(2alpha))/(sin(2alpha)cos(4alpha))#

# = (2cos(2alpha))/(cos(4alpha))#

Evaluating the limit as #alpha rarr0# we get

#(2cos(0))/cos(0) = (2(1))/(1) = 2#

We can also use the standard limit #lim_(xrarr0)sin(x)/x=1#. This can be inverted to say that #lim_(xrarr0)x/sin(x)=1#.

We can rearrange the given limit to make these appear:

#lim_(alphararr0)tan(4alpha)/sin(2alpha)=lim_(alphararr0)sin(4alpha)1/cos(4alpha)1/sin(2alpha)#

Multiplying by #alpha/alpha#:

#=lim_(alphararr0)sin(4alpha)/alpha1/cos(4alpha)alpha/sin(2alpha)#

Multiplying by #(2(2))/4#:

#=2(lim_(alphararr0)sin(4alpha)/(4alpha)1/cos(4alpha)(2alpha)/sin(2alpha))#

We can split up the limit. We also see that #lim_(alphararr0)sin(4alpha)/(4alpha)# and #lim_(alphararr0)(2alpha)/sin(2alpha)# are in the form #lim_(xrarr0)sin(x)/x=1#.

#=2*lim_(alphararr0)(2alpha)/sin(2alpha)*lim_(alphararr0)1/cos(4alpha)*lim_(alphararr0)(2alpha)/sin(2alpha)#

#=2*1*1/cos(0)*1#

#=2#