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