Question

What is the limit of `[(5^x)-1]/x` as `x` approaches `0`?

Answer 1

`lim_(x->0) (5^x-1)/x = ln 5`

Explanation:

Use:

`e^t = 1+t+t^2/2+t^3/6+t^4/24+... = sum_(k=0)^oo t^k/(k!)`

Note that:

`5^x = (e^(ln 5))^x = e^(x ln 5)`

Let:

`t = x ln 5`

Then:

`(5^x-1)/x = ((e^t-1)/t)ln 5`

Now:

`(e^t-1)/t = ((sum_(k=0)^oo t^k/(k!)) - 1)/t`

`color(white)((e^t-1)/t) = ((sum_(k=0)^oo t^k/(k!)) - 1)/t`

`color(white)((e^t-1)/t) = (sum_(k=1)^oo t^k/(k!))/t`

`color(white)((e^t-1)/t) = sum_(k=1)^oo t^(k-1)/(k!)`

`color(white)((e^t-1)/t) = sum_(k=0)^oo t^k/((k+1)!)`

`color(white)((e^t-1)/t) = 1+sum_(k=1)^oo t^k/((k+1)!)`

So:

`lim_(t->0) (e^t-1)/t = 1+lim_(t->0) (sum_(k=1)^oo t^k/((k+1)!))= 1+0 = 1`

and:

`lim_(x->0) (5^x-1)/x = lim_(t->0) ((e^t-1)/t) ln 5 = ln 5`

Answer 2

See below.

Explanation:

From the differential definition

`f'(x) = lim_(h->0)(f(x+h)-f(x))/h`

making `f(x) = 5^x = e^(xlog5)` we have

`lim_(h->0)(f(0+h)-f(0))/h = f'(0) = log5`