Question

若函数f(x)=a(x+1),x<=0,(1+2x)^1/x,x>0.在x=0处连续，则a=___【高数题，求大佬详解】

题如图下(插本题）拜托了

Answer 1

右极限 lim<x→0>(1+2x)^(1/x) = lim<x→0>[(1+2x)^(1/2x)]^2 = e^2
左极限 = a = f(0) = e^2, a = e^2