Show that [1+(1/x)]^x is an increasing function of x.?

Answer 1

For x a natural number, ie, positive integer (1, 2, 3 ...)

that's the sequence for e, Euler's number. The base of the natural log.

You can program your TI-83 to show a sequence and plot the values to see it converge. You'll also see the values in the table.

Induction might be the best way to show that it's monotonic.

Either that, or show that the ratio of consequtive terms is greater than 1. I'd recommend the ratio test.

S(n+1)/S(n) = [1+1/(n+1)]^(n+1)/[1+1/n]^n =
yuk, maybe that's not the best.

Answer 2

You can only do that by finding the derivative. Logarithmic differentiation will give it to you.

If f(x)= (1 + 1/x)^x, then

log(f) = x log(1 + 1/x)

f'/f = log(1 + 1/x) + [x/(1+1/x)]*(-1/x^2)

f' = (that stuff in the last line)*f

Now you just show that f' is positive. Finding f'' and showing it is positive and never zero is probably the hard way to do that, but the other way is to manipulate f' into a form that is obviously positive for all x, and I don't see an easy way to do that.

Answer 3

♠ Sarah, both John and Modulo started good, pity they’ve pooped out and haven’t brought us to a logical end. The lim{(x→+∞) of (1 + 1/x)^x} =e, also mentioned as 2nd remarkable limit; I take John’s start.
♣ so let x>=1; see John; y'/y = ln(1+1/x) +(x/(1+1/x))*(-1/x^2) = ln(1+1/x) -1/(1+x); thus y’ =y*{ln(1+1/x) -1/(1+x)};
now since
1) y’(1) = (1+1)^1 *{ln(1+1/1) –1/(1+1)} = 2ln2 -1 = 0.385 >0; {here we might take any fixed x>=1};
2) both functions ln(1+1/x) and 1/(1+x) are monotonous;
3) The lim{(x→+∞) of {ln(1+1/x) -1/(1+x)} =0, while y(x) → e;
◙ thence y’(x)>0 for any x>=1, hence y(x) is monotonous increasing function with horizontal asymptote y=e;