Compute the limit $\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n$

Question

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?

Answer 1

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.

Answer 2

$$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{\sqrt[n]{a}-1}{2}\right)^{\frac{2}{\sqrt[n]{a}-1}\cdot\frac{1}{2}\cdot\frac{\sqrt[n]a-1}{\frac{1}{n}}}=e^{\frac{1}{2}\ln{a}}=\sqrt{a}$$

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

Answer 3

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 $\frac{1}{2}\log a$.

Answer 4

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).

Answer 5

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 $$ \lim_{t\to0}\frac{\ln(1+a^t)-\ln2}{t}= (\ln a)\lim_{u\to0}\frac{\ln(1+u)}{\ln(1+2u)}= \frac{1}{2}(\ln a)\lim_{u\to0}\frac{\ln(1+u)}{u}\frac{2u}{\ln(1+2u)} $$

Answer 6

$$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n =\lim_{n\rightarrow\infty}\exp\left(n\ln \left(1+\frac{\sqrt[n]{a}-1}{2}\right)\right) \\=\lim_{n\rightarrow\infty}\exp\left(n\frac{\sqrt[n]{a}-1}{2}\frac{\ln \left(1+\frac{\sqrt[n]{a}-1}{2}\right)}{\frac{\sqrt[n]{a}-1}{2}}\right) \\=\lim_{n\rightarrow\infty}\exp\left(n\frac{\sqrt[n]{a}-1}{2}\right)\\=\lim_{n\rightarrow\infty}\exp\left(\frac{1}{2}\frac{a^{\frac{1}{n}}-1}{\frac{1}{n}}\right) \\=\lim_{h\rightarrow0}\exp\left(\frac{1}{2}\ln a\right) =\sqrt a$$

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, $$\color{blue}{\ln a =(\ln a\cdot a^h)|_{h=0}= \frac{d}{dh}(a^h)|_{h=0} = \lim_{h\rightarrow0}\frac{a^{h}-1}{h}} $$

Answer 7

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

Answer 8

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

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