Evaluate the limit? : #lim_(x rarr 0) ( tanx-x ) / (x-sinx) #
1 Answer
# lim_(x rarr 0) ( tanx-x ) / (x-sinx) = 2 #
Explanation:
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
# 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 #
