# Limits for sequences and the $e$ method

After doing a lot of homework, I've realized that I don't really get how to use the $\left(1+\frac{1}{n}\right)^n\to e$ method, which shows in how I can't solve indeterminate limits of the type $1^\infty$. For example:

$$\lim_{n\to\infty}{\left(\frac{1}{2}+\frac{2}{n}\right)^n}$$

Then I would do as follows:

$$\lim_{n\to\infty}{\left(\frac{1}{2}+\frac{1}{2}+\frac{2}{n}-\frac{1}{2}\right)^n}$$

$$\lim_{n\to\infty}{\left(1+\frac{4-n}{2n}\right)^n}$$

$$\lim_{n\to\infty}{\left(1+\frac{1}{\frac{2n}{4-n}}\right)^n}$$

And, from here on, I would not know what to do. Thus, the question is:

How am I supposed to work with the exponents?

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You do not need to use that limit. For $n>8$, note that $({1\over2}+{2\over n})^n< (3/4)^n$. Use the Squeeze Theorem. – David Mitra Sep 13 '12 at 1:15

Write $$\left(\frac{1}{2} + \frac{2}{n}\right)^n = \frac{1}{2^n} \left(1+\frac{4}{n}\right)^n$$ Taking limits, we get $$\lim_{n \rightarrow \infty} \left(\frac{1}{2} + \frac{2}{n} \right)^n = 0 \cdot e^4 = 0$$

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$\big(\frac{1}{2}+\frac{2}{n}\big)^n=\big(\frac{1}{2}(1+\frac{2/n}{1/2})\big)^n$‌​. – S. Snape Sep 13 '12 at 1:32

Another approach:

If $\lim_{x\to{+\infty}} f(x)^{g(x)}$ is as $costant^{+\infty}$ which is indeterminate form then: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim_{x\to +\infty}\big(f(x)-1\big)g(x)}$$

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Hello!! How are you? I've missed you today...maybe you'll read this when you awaken, or soon after! :-) – amWhy Mar 18 '13 at 0:55
@amWhy: Hi! Amy. ;-) – S. Snape Mar 18 '13 at 2:40

Here's a trick:

$$\lim_{n\rightarrow \infty}\left(\frac{1}{2}+\frac{2}{n}\right)^n =\lim_{n\rightarrow \infty}e^{n\ln\left(\frac{1}{2}+\frac{2}{n}\right)}$$ which is $\,0\,$ since the limit of $$n\ln\left(\frac{1}{2} + \frac{2}{n}\right)$$ as n tends to infinity is $-\infty$.

Note, you can also use the trick above to prove that $$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n = e$$

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Your final limit, $$\lim_{n\to\infty}{\left(1+\frac{1}{\frac{2n}{4-n}}\right)^n}\;,$$ does not have the right form for you to apply the fact that $\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$, because the fraction $$\frac{1}{\frac{2n}{4-n}}$$ does not tend to $0$ as $n\to\infty$. Neither does the fraction $\dfrac{4-n}{2n}$ from the preceding step. Thus, either you need a different rearrangement altogether, or you need to use a different approach. As others have already suggested, a different approach does the job nicely.

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