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4 March, 15:12

Lim as x approaches infinity (1+3/x) ^ (5x)

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  1. 4 March, 18:04
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    1 because as x approach infinity the 3/x goes to zero which just leaves 1^ (infinity) and 1 to any power is 1.

    they show that lim x-> infinity (1+a/x) ^x=e^a

    so if we have lim x-> infinity (1+3/x) ^x=e^3

    and if we have lim x-> infinity (1+3/x) ^ (5x) = lim x->infinity [ (1+3/x) ^x]^5

    (the inside goes to e^3) = e^[3 (5) ]=e^15
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