Limit of Function

The entire limit can be written as :-

$\displaystyle \lim_{n\to\infty} \frac{a^{\frac{1}{n}} - a^{\frac{1}{n+1}}}{\frac{1}{n^{2}}}$ ( $\frac{0}{0} form)$

So By L'Hospital Rule :-

$\displaystyle \lim_{n\to\infty} (\ln(a)).\frac{(\frac{1}{(n+1)^{2}} - \frac{1}{n^2})}{\frac{-2}{n^3}}$

(Note that I have made the approximation that $\large a^{\frac{1}{n}}$ tends to 1 as n tends to $\large \infty$ )

So rearranging we see that it is :-

$\displaystyle \lim_{n\to\infty} (\ln(a)).\frac{(2n+1).n^{3}}{2n^{2}(n+1)^{2}}$ .

which tends to $1$ .

Let $f(x) = -a^{1/x}$ . We are considering the behavior of the product $n^2(f(n+1)-f(n))$ as $n \rightarrow \infty$ .

Apply the mean value theorem to $(f(n+1)-f(n))$ : for every value of $n$ , there exists a value $x_n \in (n,n+1)$ such that $f'(x_n) = (f(n+1)-f(n))$ . (I'm implicitly using the fact that the length of the interval is 1 to ignore the denominator of the MVT.)

Note that since $f(x) = -\exp(\frac{\ln a}{x})$ , $f'(x) = - \exp(\frac{\ln a}{x}) (-\frac{\ln a}{x^2}) = -f(x) \frac{\ln a}{x^2}$ .

Thus we are interested in the limit $\lim_{n \rightarrow \infty} n^2 f'(x_n) = \lim_{n \rightarrow \infty} n^2 (-f(x_n)) \frac{\ln a}{x_n^2}$ .

Now it should be clear that $\lim_{n \rightarrow \infty} f(x_n) = -1$ , since $\lim_{x \rightarrow \infty} (1/x) = 0$ . Thus $\lim_{n \rightarrow \infty} -f(x_n) = 1.$

Also, note that, since $n < x_n < n+1$ for all $n$ , then $\lim_{n \rightarrow \infty} \frac{n}{x_n} = 1$ .

Thus $\lim_{n \rightarrow \infty} n^2 (-f(x_n)) \frac{\ln a}{x_n^2} = \lim_{n \rightarrow \infty} \frac{n^2 \ln a}{x_n^2} = \ln a$ .

make a change of variable $n\to \dfrac{1}{n}$ , to have $\lim_{n\to 0} \dfrac{1}{n^2} \left( a^{n}-a^{\frac{n}{n+1}} \right)=\lim_{n\to 0} \dfrac{1}{n^2} \left[ \exp( n \ln(a)) -\exp\left(\frac{n}{n+1}\ln(a)\right) \right]=\lim_{n\to 0} \dfrac{1}{n^2} \left[ \exp( n \ln(a)) -\exp\left(n\ln(a)-n^2\ln(a)+O(n^3)\right) \right] \\ =\lim_{n\to 0} \dfrac{1}{n^2} \left[ 1+n\ln(a)+\dfrac{n^2 \ln^2 (a)}{2}+O(n^3) -\left(1+ \left(n\ln(a)-n^2\ln(a)\right)+\dfrac{\left(n\ln(a)-n^2\ln(a)\right)^2}{2}+O(n^3)\right)\right] \\ =\lim_{n\to 0} \dfrac{1}{n^2} \left[ n^2\ln(a) +O(n^3)\right] = \boxed{\ln(a)}$

We just have to suppose $a=1$ , then $\text{The Equation}=0$ , $\text{ln }a=0$ . So the answer is $\text{ln }a$ .

$\begin{aligned} L & = \lim_{n \to \infty} n^2 \left(\sqrt[n] a - \sqrt[n+1] a\right) \\ & = \lim_{n \to \infty} \frac {a^\frac 1n - a^\frac 1{n+1}}{\frac 1{n^2}} & \small \color{#3D99F6} \text{Let }x = \frac 1n \\ & = \lim_{x \to 0} \frac {a^x - a^\frac x{1+x}}{x^2} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies} \\ & = \lim_{x \to 0} \frac {a^x \ln a - \frac {a^\frac x{1+x}\ln a}{(1+x)^2}}{2x} & \small \color{#3D99F6} \text{After }\frac d{dx} \text{up and down, a 0/0 case again} \\ & = \lim_{x \to 0} \frac {a^x \ln^2 a - \frac {a^\frac x{1+x}\ln^2 a}{(1+x)^4} + \frac {2 a^\frac x{1+x}\ln a}{(1+x)^3}}2 & \small \color{#3D99F6} \frac d{dx} \text{up and down again} \\ & = \boxed{\ln a} \end{aligned}$

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Reference: L'Hôpital's rule

Since $\frac{1}{n}-\frac{1}{n+1} = \frac{1}{n(n+1)}$ , we have: $n^2\left ( a^{\frac{1}{n}}-a^{\frac{1}{n+1}} \right ) = {\color{#0C6AC7}\frac{{}a^{\frac{1}{n+1} + \frac{1}{n(n+1)}}-a^{\frac{1}{n+1}}}{\frac{1}{n(n+1)}} } \cdot \frac{n^2}{n(n+1)}$ Taking the limit as $n \to \infty$ , we get that the desired limit is equal to: ${\color{#0C6AC7}\frac{d}{dx}\left ( a^{x} \right )\Bigr|_{x = 0}} = \ln{(a)}\cdot a^{x}\Bigr|_{x = 0} = \ln{a}$

$\begin{aligned} n^2(a^{1/n}-a^{1/(n+1)}) &= n^2a^{1/(n+1)}(a^{1/n(n+1)}-1)\\ &=n^2a^{1/(n+1)}(e^{\ln(a)/n(n+1)}-1)\\ &\underset{n\to +\infty}{=} n^2\left[\frac{\ln(a)}{n(n+1)}+o\left(\frac{1}{n(n+1)}\right)\right]\\ &=\boxed{\ln(a)}+o(1) \end{aligned}$

Note:

• $e^u\underset{u\to 0}{=}1+u+o(u)$

• $o\left(\frac{1}{n(n+1)}\right)=o\left(\frac{1}{n^2}\right)$

• $a^{1/(n+1)}\rightarrow 1$

• $\frac{n^2}{n(n+1)}\rightarrow 1$

use Taylor's formula $\sqrt[n]{a} = e^{\frac{\ln a}{n}} = 1 + \frac{\ln a}{n} + \frac{\ln^2 a}{2n^2} + o(\frac{1}{n^3})$

so $\sqrt[n]{a} - \sqrt[n+1]{a} = \frac{\ln a}{n(n+1)} + o(\frac{1}{n^3})$

thus $\lim_{n\to \infty} n^2 (\sqrt[n]{a} - \sqrt[n+1]{a}) = \lim_{n\to \infty} \frac{n^2\ln a}{n(n+1)} = \ln a$

Just do a=1 and it is quite obvious that the limit is 0 as the nth and the (n+1)th root of 1 is 1 so their difference is 0. The only answer that would make this work is ln(a).

More of a remark, not a full answer: A cheap way to get the answer is observe that when $a = 1$ , $\sqrt[n]{a} - \sqrt[n+1]{a} = 1^{n} - 1^{n+1} = 1 - 1 = 0$ so $\lim_{n\to\infty} n^2 \cdot 0 = \lim_{n\to\infty} 0 = 0$ . But only $\ln(1) = 0$ ... Maybe consider adding $0$ as an option?

Im not sure if I’m right but If we let a = 1 and the limit = 0 as n approaches infinity so the only one of options that fits this condition is ln a

The simplest solution I found was setting a = 1. In this case the answer is always 0. The only answer that fits this is ln(a). If 0 < a < 1, the second term is always negative, and if a > 1, the second term is always positive. This gives more evidence that the only possible answer is ln(a).