Indeterminate Form

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September 28th 2009, 12:57 PM
soma

Indeterminate Form

x -> 0

What is the indeterminate form of

lim n-> infinity (1 + 7/n + x/n)^n

If I let u = (7+x)/n, the problem becomes

lim n-> infinity (1+u)^((7+x)/u)

And the answer is e^7, but I don't understand the process of obtaining the answer.
September 28th 2009, 11:10 PM
mr fantastic

Quote:

Originally Posted by soma

x -> 0

What is the indeterminate form of

lim n-> infinity (1 + 7/n + x/n)^n

If I let u = (7+x)/n, the problem becomes

lim n-> infinity (1+u)^((7+x)/u)

And the answer is e^7, but I don't understand the process of obtaining the answer.

Actually the answer is $e^{7 + x}$ .

You have $\lim_{n \rightarrow + \infty} \left( 1 + \frac{7 + x}{n}\right)^n = \lim_{n \rightarrow + \infty} \left( 1 + \frac{a}{n}\right)^n = e^a$ .

(The last one is a standard limit and should be known).
September 29th 2009, 04:59 AM
soma

for:

$\lim_{n \rightarrow + \infty} \left( 1 + \frac{a}{n}\right)^n = e^a$

How can I show the work on my homework assignment for that particular step? I know you say it's a standard limit, but I can't find any information about it online... Or is it just a matter of saying it's a standard limit?
September 29th 2009, 05:18 AM
Danny

Quote:

Originally Posted by soma

for:

$\lim_{n \rightarrow + \infty} \left( 1 + \frac{a}{n}\right)^n = e^a$

How can I show the work on my homework assignment for that particular step? I know you say it's a standard limit, but I can't find any information about it online... Or is it just a matter of saying it's a standard limit?

Here's a site

http://en.wikipedia.org/wiki/E_(mathematical_constant)

As for your problem, from

$\lim_{n \to + \infty} \left( 1 + \frac{1}{n}\right)^n = e$

set $n = \frac{m}{a}$ noting that $n \to \infty$ gives $m \to \infty$ so

$\lim_{m \to + \infty} \left( 1 + \frac{a}{m}\right)^{m/a} = e$

so

$\left(\lim_{m \to + \infty} \left( 1 + \frac{a}{m}\right)^{m/a} \right)^a = e^a$ .
September 29th 2009, 05:31 AM
soma

That's perfect. I now understand the problem completely, thanks!