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NOTE: L'Hopital's and Taylor series not allowed!

Taking the log and exponenting the entire thing I get $$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}e^{n\ln\left(\frac{1+\sqrt[n]{a}}{2}\right)}=\lim_{n\rightarrow\infty}e^{n\ln\left(\frac{1+\sqrt[n]{a}}{2}-1+1\right)\cdot\frac{\frac{1+\sqrt[n]{a}}{2}-1}{\frac{1+\sqrt[n]{a}}{2}-1}}.$$

Letting $k=\frac{1+\sqrt[n]{a}}{2}-1$ we see that the RHS can be simplified to

$$\lim_{n\rightarrow\infty}e^{nk\frac{\ln\left(k+1\right)}{k}}.$$

It follows that $k\rightarrow0$ as $n\rightarrow\infty,$ and $\ln(k+1)/k$ then tends to $1$ (standard limit). So we can write

$$\lim_{n\rightarrow\infty}(e^{k})^n=\lim_{n\rightarrow\infty}\left(\frac{e^{k}-1+1}{k}\cdot k\right)^n=\lim_{n\rightarrow\infty}\left( \frac{e^k-1}{k}\cdot k+1\right)^n=(1 \cdot0+1)^n=1.$$

The answer should be $\sqrt{a}.$ Why is my method wrong?

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marked as duplicate by Paramanand Singh calculus Oct 19 '17 at 2:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Note that you cannot generally assign a value to $ [1^\infty] $. $\endgroup$ – derpy Oct 18 '17 at 20:39
  • $\begingroup$ I wonder why you took a wrong turn when you had almost arrived at the answer. Since $(1/k)\log(1+k)\to 1$ you are left with $(1/2)n(\sqrt[n]{a}-1)$ in the exponent and we use the famous limit $n(\sqrt[n] {a} - 1)\to \log a$ to get the answer as $e^{(\log a) /2}=\sqrt{a}$. $\endgroup$ – Paramanand Singh Oct 19 '17 at 2:56
  • $\begingroup$ If you know that $(1+(x/n))^{n}\to e^{x} $ then you should be aware of its counterpart $n(\sqrt[n] {x} - 1)\to\log x$. $\endgroup$ – Paramanand Singh Oct 19 '17 at 2:57
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Letting $\frac{1}{n}=x$ and noting \begin{eqnarray} &&\lim_{n\rightarrow\infty}\ln\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}\frac{\ln\left(\frac{1+a^{\frac1n}}{2}\right)}{\frac1n}\\ &=&\lim_{x\rightarrow0}\frac{\ln\left(\frac{1+a^{x}}{2}\right)}{x}=\lim_{x\rightarrow0}\frac{\ln\left(1+\frac{a^{x}-1}{2}\right)}{x}\\ &=&\lim_{x\rightarrow0}\frac{\ln\left(1+\frac{a^{x}-1}{2}\right)}{\frac{a^{x}-1}{2}}\frac{\frac{a^{x}-1}{2}}{x}\\ &=&\frac{1}{2}\ln a, \end{eqnarray} one has $$ \lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\sqrt a. $$ Here $$ \lim_{x\to0}\frac{\ln(1+x)}{x}=1, \lim_{x\to0}\frac{a^x-1}{x}=\ln a$$ are used.

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  • $\begingroup$ I donät understand how the limit equates to $\frac{1}{2}\ln{a}$ when $x\rightarrow{0}$. On the right factor you just get 0 in the denominator and $a^x\rightarrow 1$. So one would be left with $((1-1)/2)/0.$ $\endgroup$ – Parseval Oct 18 '17 at 21:08
  • $\begingroup$ @Parseval, see the update. $\endgroup$ – xpaul Oct 18 '17 at 21:14
  • $\begingroup$ Ok now I get it. Never knew that standard limit of $(a^x-1)/x.$ Thanks! $\endgroup$ – Parseval Oct 18 '17 at 21:18
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$$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{\sqrt[n]{a}-1}{2}\right)^{\frac{2}{\sqrt[n]{a}-1}\cdot\frac{1}{2}\cdot\frac{\sqrt[n]a-1}{\frac{1}{n}}}=e^{\frac{1}{2}\ln{a}}=\sqrt{a}$$

I used the following.

We know that $\frac{\sqrt[n]{a}-1}{2}\rightarrow0$ for $n\rightarrow+\infty$.

Thus, if $\frac{\sqrt[n]{a}-1}{2}=y$ then $(1+y)^{\frac{1}{y}}\rightarrow e$.

Also, we know that $\lim\limits_{x\rightarrow0}\frac{e^x-1}{x}=1$.

Thus, $$\lim_{n\rightarrow+\infty}\frac{\sqrt[n]a-1}{\frac{1}{n}}=\lim_{x\rightarrow0}\frac{a^x-1}{x}=\ln{a}\lim_{x\rightarrow0}\frac{e^{x\ln{a}}-1}{x\ln{a}}=\ln{a}.$$

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    $\begingroup$ Michael plzz you did it again... I have three questions: 1) Would you mind editing that amazing post and fill inte some explanations and details? 2) what education do you have? 3) what do you work with? Reason for asking the two latter questions is because I can't understand how you can look at something for 30 secs and come up with a solution in one line. I've struggled with this problem all day! Any general tips/tricks/suggestions? $\endgroup$ – Parseval Oct 18 '17 at 20:48
  • $\begingroup$ @Parseval I added something. See now. $\endgroup$ – Michael Rozenberg Oct 18 '17 at 21:14
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As a general rule, if $a_n$ is a sequence such that $\lim_{n\to\infty} n(a_n-1)=b$ then $\lim_{n\to\infty} a_n^{n}=e^b$.

I've proved the case $b=0$ in this answer.

The case of general $b$ is a corollary, letting $a_n'=\frac{a_n}{1+b/n}$. Then:

$$n(a_n'-1)=\frac{n(a_n-1)-b}{1+b/n}\to 0.$$

So the fact that $a_n'^n\to 1$ means $a_n^n\to e^b$.

So you need to find $\lim_{n\to\infty} n\cdot \frac{\sqrt[n]{a}-1}{2}.$ This is half the derivative of $a^x$ at $x=0$, which is $\frac{1}{2}\log a$.

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For

The answer should be $\sqrt{a}$. Why is my method wrong?

Your mistake is at $$\lim_{n \to \infty} \exp\left(nk\dfrac{\ln(k+1)}{k}\right)= \lim_{n \to \infty} \exp\left(nk\right)$$

Here you used $\lim_{x \to 0} \dfrac{\ln(x+ 1)}{x} = 1$. If we write the intermediate step that you missed then

$$\lim_{n \to \infty} \exp\left(nk\dfrac{\ln(k+1)}{k}\right)=\exp\left({\lim_{n \to \infty} nk} \lim_{n \to \infty} \dfrac{\ln(k+1)}{k}\right) = \lim_{n \to \infty} \exp\left(nk\right)$$

$\lim_{n \to \infty} nk$ is a limit of form $0 \cdot \infty$ which is an indeterminate form, so you can't use product of limits is limit of product (one of the limits in the product is not definied).

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  • $\begingroup$ Do you agree with $k --> 0 n--> \infty$ ? $\endgroup$ – Isham Oct 18 '17 at 21:15
  • $\begingroup$ @Isham Yes I do. $\endgroup$ – A---B Oct 18 '17 at 21:18
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$$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n =\lim_{n\rightarrow\infty}\exp\left(n\ln \left(1+\frac{\sqrt[n]{a}-1}{2}\right)\right) \\=\lim_{n\rightarrow\infty}\exp\left(n\frac{\sqrt[n]{a}-1}{2}\frac{\ln \left(1+\frac{\sqrt[n]{a}-1}{2}\right)}{\frac{\sqrt[n]{a}-1}{2}}\right) \\=\lim_{n\rightarrow\infty}\exp\left(n\frac{\sqrt[n]{a}-1}{2}\right)\\=\lim_{n\rightarrow\infty}\exp\left(\frac{1}{2}\frac{a^{\frac{1}{n}}-1}{\frac{1}{n}}\right) \\=\lim_{h\rightarrow0}\exp\left(\frac{1}{2}\ln a\right) =\sqrt a$$

Given that $$\color{red}{\lim_{n\to \infty}\frac{\ln \left(1+\frac{\sqrt[n]{a}-1}{2}\right)}{\frac{\sqrt[n]{a}-1}{2}} = 1~~~since ~~\lim_{X\to 0}\frac{\ln(X+1)}{X} = 1}$$ and by Hospital rule, $$\color{blue}{\ln a =(\ln a\cdot a^h)|_{h=0}= \frac{d}{dh}(a^h)|_{h=0} = \lim_{h\rightarrow0}\frac{a^{h}-1}{h}} $$

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Here is a purely algebraic proof, without any appeal to the exponential or logarithmic function.

It suffices to show that the limit is $\sqrt a$ when $a\ge1$, since

$$\left(1+\sqrt[n]{1/a}\over2\right)^n=\left(\sqrt[n]a+1\over2\sqrt[n]a\right)^n={1\over a}\left(1+\sqrt[n]a\over2\right)^n$$

Observe first that (when $a\ge1$)

$$\left(1+\sqrt[n]a\over2\right)^n\le a$$

which becomes clear by rewriting the inequality as $1+\sqrt[n]a\le2\sqrt[n]a$.

Now

$$\begin{align} \left(1+\sqrt[n]a\over2\right)^n-\sqrt a &=\left(1+\sqrt[n]a\over2\right)^n-\left(\sqrt[2n]a\right)^n\\ &=\left({1+\sqrt[n]a\over2}-\sqrt[2n]a\right)\left(\left(1+\sqrt[n]a\over2\right)^{n-1}+\cdots+\left(\sqrt[2n]a\right)^{n-1} \right)\\ &={1\over2}(\sqrt[2n]a-1)^2\left(\left(1+\sqrt[n]a\over2\right)^{n-1}+\cdots+\left(\sqrt[2n]a\right)^{n-1} \right) \end{align}$$

The general term in the "$\cdots$" is

$$\left(1+\sqrt[n]a\over2\right)^k(\sqrt[2n]a)^{(n-1)-k}=(\sqrt[2n]a)^{n-1}\left(1+\sqrt[n]a\over2\sqrt[2n]a\right)^k\le\sqrt a\left(1+\sqrt[n]a\over2\right)^n\le a^{3/2}$$

Since there are $n$ such terms in all (and they're all clearly positive), we have

$$0\le\left(1+\sqrt[n]a\over2\right)^n-\sqrt a\le{1\over2}(\sqrt[2n]a-1)^2(na^{3/2})$$

Finally, from the binomial theorem, we have

$$\left(1+{a\over2n}\right)^{2n}=1+a+\cdots\gt a$$

so that

$$0\le\sqrt[2n]a-1\lt{a\over2n}$$

This gives us

$$0\le\left(1+\sqrt[n]a\over2\right)^n-\sqrt a\le{1\over2}\left(a\over2n\right)^2(na^{3/2})={a^{7/2}\over8n}$$

The Squeeze Theorem now tells us

$$\lim_{n\to\infty}\left(1+\sqrt[n]a\over2\right)^n=\sqrt a$$

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If you are able to prove that $$ \lim_{t\to0}\frac{\ln(1+a^t)-\ln2}{t}=l $$ then your limit is $e^l$. Indeed, if $f(t)=\frac{\ln(1+a^t)}{t}$, then $$ \left(\frac{1+\sqrt[n]{a}}{2}\right)^{\!n}=e^{f(1/n)} $$ The limit to be computed is just the derivative of $g(t)=\ln(1+a^t)$ at $0$; since $g'(t)=(a^t\ln a)/(1+a^t)$, the limit is $g'(0)=\frac{1}{2}\ln a=\ln\sqrt{a}$.

If you are not allowed to use derivatives, then you can do it indirectly; set $(1+a^t)/2=1+u$, so $a^t=1+2u$ and $t=\ln(1+2u)/\ln a$, so $$ \lim_{t\to0}\frac{\ln(1+a^t)-\ln2}{t}= (\ln a)\lim_{u\to0}\frac{\ln(1+u)}{\ln(1+2u)}= \frac{1}{2}(\ln a)\lim_{u\to0}\frac{\ln(1+u)}{u}\frac{2u}{\ln(1+2u)} $$

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Since $(a^h)'= \ln a .a^h$ and $\ln\left(\frac{1+a^{0}}{2}\right)=0$, By LHOPITAL RULE we have

$$ \lim_{h\rightarrow0}\frac{\ln\left(\frac{1+a^{h}}{2}\right)}{h}=\lim_{h\rightarrow0}\frac12\frac{ a^h\ln a}{\frac{1+a^{h}}{2}}=\frac12\ln a,$$

Hence , letting $\frac{1}{n}=h$ one has $$ \lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}\exp\left(\frac{\ln\left(\frac{ 1+a^{\frac{1}{n}}}{2}\right)}{\frac{1}{n}}\right) =\sqrt a$$

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