What is the limit of [(5^x)-1]/x(5x)−1x as xx approaches 00?
2 Answers
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
See below.
Explanation:
From the differential definition
making
