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# what does (1+r/m)^m tend to as m tends to infinity? Thanks!?

I think the last answer is right but can anyone explain why

(1+1/(m/r))^r(m/r)

is e^r?

Thanks!

### 3 Answers

- Anonymous1 decade agoFavorite Answer
(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

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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

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- 1 decade ago
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.

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- Anonymous1 decade ago
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

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