Using the squeeze theorem: lim x--> 0+ (e^x -1) * 3^(2cos(π/x))

Question

lim

x->0⁺   (e^x-1) * 3^2cos(π/x)

Answer 1

the limit of a product is the product of the limits of the two factors.  (limits below are as x approaches 0⁺)

 

lim(e^x-1) = 1=1 = 0.     lim cos(n/x) = cos 0 = 1.       lim 3^2cos0 = 3²⁽¹⁾ = 9.  

Therefore overall limit that you seek is 0(9) = 0.