Which of the following functions grow faster then e^x as x approaches infinity?
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3^x because it is e^(xln 3) while e^x is just e^(x*1) and ln 3 > 1.
Originally Posted by ThePerfectHacker 3^x because it is e^(xln 3) while e^x is just e^(x*1) and ln 3 > 1. Another reason as to why...
Same exponent, but the left side has a larger base hence a larger outcome, and to connect this to TPH's explanation.. <- this follows because at zero the functions are equal
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