Evaluate the limit? : #lim_(x rarr 0) ( tanx-x ) / (x-sinx) #

Question

Evaluate the limit? : #lim_(x rarr 0)   ( tanx-x ) /  (x-sinx)  #

Emilia Aguilar · Accepted Answer

# lim_(x rarr 0) ( tanx-x ) / (x-sinx) = 2 #

We want to find: # L = lim_(x rarr 0) ( tanx-x ) / (x-sinx) # Method 1 : Graphically graph{( tanx-x ) / (x-sinx) [-8.594, 9.18, -1.39, 7.494]} Although far from conclusive, it appears that: # L = 2 # Method 2 : L'Hôpital's rule The limit is of an indeterminate form #0/0#, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exists then: # lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x)) # And so applying L'Hôpital's rule we get: # L = lim_(x rarr 0) (d/dx ( tanx-x )) / (d/dx (x-sinx)) # # \ \ \ = lim_(x rarr 0) ( sec^2x-1 ) / (1-cosx) # This limits is also of an indeterminate form, so we can apply L'Hôpital's again to get: # L = lim_(x rarr 0) (d/dx ( sec^2x-1 ) ) / (d/dx (1-cosx) ) # # \ \ \ = lim_(x rarr 0) ( 2secx(secxtanx) ) / (sinx) # # \ \ \ = lim_(x rarr 0) ( 2sec^2x(sinx/cosx) ) / (sinx) # # \ \ \ = lim_(x rarr 0) 2sec^2x(1/cosx) # # \ \ \ = 2lim_(x rarr 0) sec^3x # # \ \ \ = 2 #

Charles Anctil · Answer

undefined The limit of (tanx-x)/(x-sinx) as x approaches 0 is 1.