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$a_n = \bigg( 1 + \frac{2}{n} \bigg)^n$

$$\lim_{n\to\infty} \bigg(1+\frac{2}{\infty}\bigg)^\infty$$ so we need to use L'Hospital rule, I want to take the derivative of $\bigg( 1 + \frac{2}{n} \bigg)^n$, which I thought should be $(1 + \frac{2}{n})$, but am I doing this part wrong? Then I take the limit of the derivative and get

$$\lim_{n\to\infty} 1 + \frac{2}{n} = 1$$

But the book says the answer is $e^2$. I am really lost where $e$ comes into play

Evan Kim
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    where $e$ comes into play: $$e=\lim_\limits{n\to\infty} \left(1+\frac{1}{n}\right)^n, e^x=\lim_\limits{n\to\infty} \left(1+\frac{x}{n}\right)^n$$ (some take that as the definition of $e$) – J. W. Tanner Sep 06 '19 at 12:31
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    The sticking point here is presumably your use of L’Hospital’s rule, so you should certainly include the details of such in your question text. – Semiclassical Sep 06 '19 at 12:32
  • hmm yea I have no idea what @J.W.Tanner just did – Evan Kim Sep 06 '19 at 12:33
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    The manner in which you applied L'Hopital's rule is flawed. But unless you can describe the step-by-step process by which you attempted to apply it, there's not much to say other than that you did it wrong. – Lee Mosher Sep 06 '19 at 12:39
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    Also, in your second equation you have the limit as $x\to\infty$ rather than as $n\to\infty$. – Semiclassical Sep 06 '19 at 12:41
  • $1^\infty$ is an indeterminate form. – Bernard Sep 06 '19 at 12:42
  • I updated how I attempted to use L'Hopital's rule – Evan Kim Sep 06 '19 at 12:44
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    After you write "we get..." you just knocked off an exponent $n$. What makes you think you can do that? You just changed the whole problem. – imranfat Sep 06 '19 at 12:44

3 Answers3

1

Look up how use l'Hopital's rule. To use it on an indeterminate form $1^\infty$ as we have here, we need to take the logarithm, get an indeterminate form either $0/0$ or $\infty/\infty$, apply l'Hopital's rule to get the limit of that, then exponentiate.

\begin{align} f(n) &= \left(1+\frac{2}{n}\right)^n,\qquad\text{indeterminate form }{1^\infty} \\ \log f(n) &= n\log\left(1+\frac{2}{n}\right),\qquad\text{indeterminate form }\infty \times 0 \\ \log f(n) &= \frac{\log\left(1+2/n\right)}{1/n},\qquad\text{indeterminate form }\frac{0}{0} \\ \log f(n) &\sim \frac{\frac{d}{dn}[\log\left(1+2/n\right)]}{\frac{d}{dn}[1/n]}\\ &\text{where I wrote $\sim$ for: "has the same limit as,} \\ &\qquad\text{provided the limit on the right exists"} \\ \log f(n) &\sim \frac{-2/n^2}{(-1/n^2)(1+2/n)} = \frac{2}{1+2/n} \\ \lim\log f(n) &= 2 \\ \log \lim f(n) &= 2 \\ \lim f(n) &= e^2 . \end{align}

Question: Why is $1^\infty$ indeterminate?

Answer. Because we can get different results for limits of that form. Example

$$ \lim_{n\to\infty}\left(1+\frac{2}{n}\right)^n = e^2 \\ \lim_{n\to\infty} 1^n = 1 \\ \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2} = +\infty $$ and so on.

GEdgar
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  • Why is $1^{\infty}$ indeterminate? I was told in class the only indeterminate forms are $0/0, \infty, -\infty$ – Evan Kim Sep 06 '19 at 12:45
  • As is often the case, the use of L'Hospital's rule is overkill in this argument: If one replaces integer $n$ with $2/x$ and considers the limit of $\frac{1}{x}2\ln(1+x)$ as $x \to 0$ over real $x$, then this is just the limit definition of $\frac{d}{dx}2\ln(1+x)\Big|_{x=0}$. – Semiclassical Sep 06 '19 at 13:01
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Write $$\left (1+\frac{2}{n} \right )^n = \left (1+ \frac{1}{\left (\frac{n}{2} \right )} \right )^{2\cdot\frac{n}{2}} $$ The result follows from $$\left (1+\frac{1}{m}\right ) ^m \xrightarrow[m\to\infty]{}e $$ L'Hopital's rule unnecessary and it's not clear to me how one would use it to start with.

AlvinL
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$\lim_{x\to a} (f(x))^{g(x)}$ $\\$ Where $\lim_{x\to a}f(x) =1\, and \lim_{x\to a}g(x)\,=\infty $ $\\$ Is $ e^{\lim_{x\to a}( g(x)(f(x)-1)) } $ $\\$ Hence $ e^{\lim_{n\to \infty}({n}(\frac{2}{n}))}\, = \,e^2 $

mathsdiscussion.com
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