lim x-->∞ (1+a/x)^(bx) = ?

Question

lim x-->∞ (1+a/x)^(bx) = ?
a and b are constants
It is indeterminate (1^∞) If I am not mistaken.

I set it equal to y:
y = lim x-->∞ (1+a/x)^(bx)
and started off like:
lim x-->∞ ln y = lim x-->∞ bx*ln(1+a/x)
lim x-->∞ ln y = lim x-->∞ [ln(1+a/x)]/(bx)^-1
lhospital lead me to increasing complex versions of (∞/∞)/0.

railrule! (who was that masked person?)

Answer 1

hi

suppose
y = x/a -> x = a.y

1+a/x = 1 + 1/y

(1+a/x)^(bx)
= (1 +1/y)^(a.b.y)
= [(1 +1/y)^(y)](ab)

then :)
limit : e^(a.b)

bye

Answer 2

lim (1+a/x)^(bx) = 1
x→∞

Run a few terms of the binomial expansion. You'll find that all of them approach 0 as x approaches ∞. That leaves only the first term, 1.

Answer 3

My answer would be 1, because something over a infinity is zero, so 1+0 is 1, and 1 to the infinity is 1. If l'hospital rule is needed then you wrote out the problem wrong.

Answer 4

You're correct in that it is 1^∞. However, I don't think that's indeterminate, since 1 raised to ANY power is 1. I'm pretty sure the answer is 1.

Answer 5

Zero

Answer 6

lim x->a million (|x|-x)/(x-a million) = 0 u = (|x| -x) = 0 for all x > 0, u' = 0, x > 0 v =(x-a million), v' = a million from L'Hopital's Rule, lim x->a million (|x|-x)/(x-a million) = lim x->a million u'(x)/v'(x) = 0/a million = 0 ...

Answer 7

= lim (1+(ab)/(bx) ) ^(bx)
= exp (ab) if b>0

Answer 8

Sorry, beyond my mathematics skill level.... Try an easier course. :-)

Answer 9

it's midnight. go away.

Answer 10

um....what!?!?!?...good luck wit that...