Question #34fda Calculus Limits Continuous Functions 1 Answer Konstantinos Michailidis Apr 15, 2017 Because #lim_(x->0^-)f(x)!=lim_(x->0^+)f(x)# and is different from #f(0)=0#. We have that #lim_(x->0^-)[absx]/(x^2+2x)=lim_(x->0^-)[-x]/[x^2+2x]=lim_(x->0^-)-[1]/[x+2]=-1/2# and #lim_(x->0^+)[absx]/(x^2+2x)=lim_(x->0^+)[x]/[x^2+2x]=lim_(x->0^+)[1]/[x+2]=1/2# The graph of f(x) around zero is 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 1596 views around the world You can reuse this answer Creative Commons License