Evaluate the limit? # lim_(x rarr 0) (1-cos2x+tan^2x)/(xsin2x) #

1 Answer
Steve M
Sep 12, 2017

I assume that we seek:

# L = lim_(x rarr 0) (1-cos2x+tan^2x)/(xsin2x) #

This is of an indeterminate form #0/0# so we can apply L'Hôpital's rule:

# L = lim_(x rarr 0) (d/dx(1-cos2x+tan^2x))/(d/dxxsin2x) #

# \ \ = lim_(x rarr 0) ( 0+2sin2x+2tanxsec^2x )/(2xcos2x+sin2x) #

# \ \ = 2lim_(x rarr 0) ( sin2x+tanxsec^2x )/(2xcos2x+sin2x) #

Again, this is of an indeterminate form #0/0# so we can apply L'Hôpital's rule a second time:

# L = 2lim_(x rarr 0) ( d/dx(sin2x+tanxsec^2x) )/( d/dx( 2xcos2x+sin2x) ) #

# \ \ = 2lim_(x rarr 0) ( 2cos2x + 2sec^2x tan^2 x +sec^4x )/( 2cos2x - 2xsin2x+ 2cos2x ) #

# \ \ = 2 ( 2 + 0 + 1 )/( 2 - 0 + 2 ) #

# \ \ = 2 ( 3 )/( 4 ) #

# \ \ = 3/2 #

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