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#

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