Yahoo Answers: Answers and Comments for What does (1+r/m)^m tend to as m tends to infinity? Thanks!? [Mathematics]
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From Anonymous
enUS
Sat, 22 Nov 2008 02:46:20 +0000
3
Yahoo Answers: Answers and Comments for What does (1+r/m)^m tend to as m tends to infinity? Thanks!? [Mathematics]
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From Anonymous: (1+r/m)^m
(1+1/(m/r))^m
(1+1/(m/r))^mr/r
(1...
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Sat, 22 Nov 2008 11:10:46 +0000
(1+r/m)^m
(1+1/(m/r))^m
(1+1/(m/r))^mr/r
(1+1/(m/r))^r(m/r)
as m > infin, so does m/r
(1+1/(m/r))^r(m/r)
This is e^r
Lets evaluate the limit mathematically.
lim (1+r/m)^m
e^ln lim (1+r/m)^m
e^lim ln (1+r/m)^m
e^lim m ln(1+r/m)
e^lim ln(1+r/m) / (1/m)
e^lim d/dm ln(1+r/m) / d/dm (1/m)
e^lim r d/dm(1/m) / [(1+r/m) d/dm (1/m) ]
e^lim r / (1+r/m)
e^r
=====
Hello.
The last answer is right? No... my answer is right. In fact, my answer is the EXACT SAME answer as the last person. The last answerer repeated exactly what I stated. So dont you dare give that person credit for what I already told you.
The limit as h → ∞
lim_{h → ∞} (1 + 1/h)^h = e
This is the definition of 'e'
And the above proof in the first half of this post explains why your limit is e^r

From chon_kah@ymail.com: 1. As m tends to infinity, the term r/m tends ...
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Sat, 22 Nov 2008 10:54:50 +0000
1. As m tends to infinity, the term r/m tends to 0 since m becomes larger and larger. Hence the term r/m gets smaller and smaller until it tends to 0.
So (1+ r/m) tends to 1 when m tends to inifinity. And we all know that 1 raised to the power of any value is still 1.

From Anonymous: As m tends to infinity, r/m tends to 0 and so ...
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Sat, 22 Nov 2008 10:52:41 +0000
As m tends to infinity, r/m tends to 0 and so the bit in the brackets tends to1.
As m tends to infinity therefore the equation tends to 1^m = 1