limit of $ \lim_{x \rightarrow 1} \frac{ f_{n} (x)- f_{n-1} (x)}{ (1-x)^{n} }=? $

Question

Let

$$ f_{n} (x)= x^{ x^{\scriptstyle\cdot^{\scriptstyle\cdot^{\scriptstyle\cdot^{\scriptstyle x}}}}}$$

Then

$$ \lim_{x \rightarrow 1} \frac{ f_{n} (x)- f_{n-1} (x)}{ (1-x)^{n} }={?} $$

My try:

\begin{align} \lim_{x \rightarrow 1} \frac{f_n-f_{n-1}}{(1-x)^{n}} &=\lim_{x \rightarrow 1}{ \frac{ e^{\ln f_{n} (x) } - e^{\ln f_{n-1} (x) } }{(1-x)^{n}} } \\[6px] &=\lim_{x \rightarrow 1}{ \frac{ e^{f_{n-1} (x)\ln x } - e^{f_{n-2} (x)\ln x } }{(1-x)^{n}} } \\[6px] &=\lim_{x \rightarrow 1} \frac{\ln x(f_{n-1} (x)-f_{n-2} (x))}{ (1-x)^{n} } \end{align}

Now?

You have almost done it. If the limit is $a_n$ then you have shown that $a_{n} =-a_{n-1}=(-1)^{n-2}a_{2}=(-1)^{n}$. — Paramanand Singh, May 19 '17 at 17:31
I think you have solved the question almost. Just use the limit $\lim_{x\to 1}\dfrac{\log x} {1-x}=-1$ and then you get the desired relation $a_{n} =-a_{n-1}$. I am on a smartphone and it is difficult to type complete answers. — Paramanand Singh, May 19 '17 at 17:38

score 5 · Answer 1 · answered May 20 '17 at 12:59

5

You are extremely close to the solution. By setting $L_n=\lim_{x\to 1}\frac{f_n(x)-f_{n-1}(x)}{(x-1)^n}$ we have

$$ L_1 = \lim_{x\to 1}\frac{x-1}{x-1}=1,\quad L_2=\lim_{x\to 1}\frac{x^x-x}{(x-1)^2}= 1$$ and $$\begin{eqnarray*} L_n = \lim_{x\to 1}\frac{f_n(x)-f_{n-1}(x)}{(x-1)^n}&=&\lim_{x\to 1}\frac{\exp(f_{n-1}(x)\log x)-\exp(f_{n-2}(x)\log x)}{(x-1)^n}\\&=& \lim_{x\to 1}\frac{\log x}{x-1}\cdot \frac{f_{n-1}(x)-f_{n-2}(x)}{(x-1)^{n-1}}\\&\stackrel{d.H.}{=}&\lim_{x\to 1}\frac{f_{n-1}(x)-f_{n-2}(x)}{(x-1)^{n-1}}=L_{n-1}\end{eqnarray*}$$ hence it follows that $L_n=\color{red}{1}$ for any $n\geq 1$.

answered May 20 '17 at 12:59

Jack D'Aurizio

353,855

Perhaps a typo. The denominator is $(1-x)^{n}$ and this leads to $(-1)^{n}$ as answer. – Paramanand Singh May 20 '17 at 14:29
@ParamanandSingh: not a typo, I dealt with such form on purpose (for simplicity). Clearly the answer to the original question depends on a adjusted sign, $(-1)^n$. – Jack D'Aurizio May 20 '17 at 16:55

limit of $ \lim_{x \rightarrow 1} \frac{ f_{n} (x)- f_{n-1} (x)}{ (1-x)^{n} }=? $

1 Answers1