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

2 Answers
George C.
Nov 5, 2017

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

Cesareo R.
Nov 5, 2017

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

Related questions
Impact of this question
7293 views around the world
You can reuse this answer
Creative Commons License