1.

Using the e-δ definition prove th( lim (2x + 3) 3x-> 0

Answer»

lim (2x + 3) = 3x-->

Let Consider ε > 0 , we will show existence of delta| 2x + 3 - 3 | < ε

| 2x - 0 | < ε

| x - 0 | < ε/2

Choose δ = ε/2implies

| x - 0 | < δ, so we have shown that for every epsilon > 0 there exist a delta .



Discussion

No Comment Found