verify an identityVerify if limit of (a^x-1)/x, is lna when x tend to 0.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the following limit, such that:

`lim_(x->0) (a^x - 1)/x = (a^0 - 1)/0 => lim_(x->0) (a^x - 1)/x = 0/0`

The indetermination of type `0/0` requests the use of l'Hospital's theorem, such that:

`lim_(x->0) (a^x - 1)/x = lim_(x->0) ((a^x - 1)')/(x')`

`lim_(x->0) ((a^x - 1)')/(x') = lim_(x->0) (a^x*ln a)/1`

`lim_(x->0) (a^x*ln a) = a^0*ln a => lim_(x->0) (a^x*ln a) = ln a`

Hence, evaluating the limit `lim_(x->0) (a^x - 1)/x` using l'Hospital's theorem, yields `lim_(x->0) (a^x - 1)/x = ln a` .

 

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll put a^x - 1 = u => a^x = 1 + u

We'll take logarithms both sides:

ln a^x = ln(1+u)

We'll use the power property of logarithms:

x lna = ln(1+u)

x = ln(1+u)/lna

We'll solve the limit: lim (a^x - 1)/x = lim u/[ln(1+u)/lna] lim u/[ln(1+u)/lna] = lim lna/(1/u)*ln(1+u)
lim lna/(1/u)*ln(1+u) = ln a*lim 1/ln(1+u)^(1/u) ln a*lim 1/ln(1+u)^(1/u) = ln a/limln(1+u)^(1/u) ln a/limln(1+u)^(1/u) = lna/ln lim (1+u)^(1/u) lna/ln lim (1+u)^(1/u) = lna/ln e = lna/1 = ln a The limit of the function, for x->0, is (a^x - 1)/x = lna. Therefore, lim (a^x-1)/x = ln a.

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