#lim_(x->0)(tan x - x)/x^3 = # ? Calculus Limits Continuous Functions 1 Answer Cesareo R. Mar 1, 2017 #-1/6# Explanation: We know that #sin x = x-x^3/(3!)+x^5/(5!)- cdots = sum_(k=0)^oo(-1)^k x^(2k+1)/((2k+1)!)# and also #sin x -x = -x^3/(3!)+x^5/(5!)- cdots# Now #(tan x - x)/x^3 = (sinx-x cos x)/(x^3 cos x)# and for small values on #x# we have #(tan x - x)/x^3 approx (sin x- x)/x^3# because then #cos x approx 1# Putting all together #lim_(x->0)(tan x - x)/x^3 = lim_(x->0)(sin x- x)/x^3 = lim_(x->0)(-x^3/(3!)+x^5/(5!)- cdots)/x^3 = -1/(3!) = -1/6# Answer link Related questions What are continuous functions? What facts about continuous functions should be proved? How do you use continuity to evaluate the limit #sin(x+sinx)# as x approaches pi? How do you find values of x for which the function #g(x) = (sin(x^20+5) )^{1/3}# is continuous? How do you find values of x where the function #f(x)=sqrt(x^2 - 2x)# is continuous? How do you use continuity to evaluate the limit sin(x+sinx)? Given two graphs of piecewise functions f(x) and g(x), how do you know whether f[g(x)] and... How do you find the interval notation to prove #f(x)= x/(sqrt(1-x^2))# is continuous? How do you use continuity to evaluate the limit #(e^(x^2) - e^(-y^2)) / (x + y)# as #(xy)#... How do you show that the function #f(x)=1-sqrt(1-x^2)# is continuous on the interval [-1,1]? See all questions in Continuous Functions Impact of this question 1517 views around the world You can reuse this answer Creative Commons License