A limit question involving power of positive numbers

Question

I'm trying compute the following limit:

$$\lim_{t\to0}\left(\frac{1}{t+1}\cdot\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t},\quad b>a>0.$$

I know $\displaystyle\lim_{t\to0}(1+t)^{1/t}=e$ and here I've found $\displaystyle\lim_{t\to0}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}=\frac{a^{\frac{a}{a-b}}}{b^{\frac{b}{a-b}}}$.

Hence the answer is $\displaystyle\frac{1}{e}\cdot\frac{a^{\frac{a}{a-b}}}{b^{\frac{b}{a-b}}}$. How can I get that answer?

A classmate of mine had a good idea, but we are stuck. The idea is take $f(x)=x^{t+1}$, by the mean value theorem exists $c\in(a,b)$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}.$$ But $$\frac{f(b)-f(a)}{b-a}=\frac{b^{t+1}-a^{t+1}}{b-a}\quad\mbox{and}\quad f'(c)=(t+1)c^t.$$ Hence, since $c>0$ $$\left(\frac{1}{t+1}\cdot\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}=(c^t)^{1/t}=c.$$ Thus $$\lim_{t\to0}\left(\frac{1}{t+1}\cdot\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}=\lim_{t\to0}c=c.$$

We are stuck here. Someone has some hint for this? Or some other idea for compute the limit above?

Thanks!

Answer 1

Let $$f(t)=\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t} =\{g(t)\}^{1/t}$$ and if $\lim_{t\to 0}f(t)=L$, then $$\begin{aligned}\log L &= \log\left\{\lim_{t \to 0}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}\right\}\\ &= \lim_{t \to 0}\,\log\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1/t}\text{ (by continuity of log)}\\ &= \lim_{t \to 0}\frac{1}{t}\log\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)\\ &= \lim_{t \to 0}\frac{1}{t}\log g(t)\\ &= \lim_{t \to 0}\frac{1}{t}\cdot\frac{\log\{1+(g(t)-1)\}}{g(t) - 1}\cdot\{g(t) - 1\}\\ &= \lim_{t \to 0}\frac{1}{t}\cdot\{g(t) - 1\}\cdot\lim_{y \to 0}\frac{\log(1+ y)}{y}\text{ (putting }y = g(t) - 1)\\ &= \lim_{t \to 0}\frac{g(t) - 1}{t}\\ &= \lim_{t \to 0}\frac{b^{t + 1} - a^{t + 1} - (b - a)}{t(b - a)}\\ &= \frac{1}{b - a}\lim_{t \to 0}\frac{b(b^{t} - 1) - a(a^{t} - 1)}{t}\\ &= \frac{1}{b - a}\lim_{t \to 0}\left(b\cdot\frac{b^{t} - 1}{t} - a\cdot\frac{a^{t} - 1}{t}\right)\\ &= \frac{1}{b - a}(b\log b - a\log a)\end{aligned}$$ It follows that $$L = (b^{b}/a^{a})^{1/(b - a)}$$ As you mention in your answer the desired limit is given by $L/e$ where $L$ is given above.

Answer 2

Why not to use $$b^{t+1}=e^{(t+1)\log(b)}$$ and use the Taylor series built at $t=0$ $$b^{t+1}=b+b t \log (b)+O\left(t^2\right)$$ Then $$\frac{b^{t+1}-a^{t+1}}{b-a}=1+\frac{ b \log (b)-a \log (a)}{b-a}t+O\left(t^2\right)$$ $$\frac{1}{t+1}\cdot\frac{b^{t+1}-a^{t+1}}{b-a}=1+\frac{ a-a \log (a)-b+b \log (b)}{b-a}t+O\left(t^2\right)$$ Now $$\log\Big(\frac{1}{t+1}\cdot\frac{b^{t+1}-a^{t+1}}{b-a}\Big)=\frac{ a-a \log (a)-b+b \log (b)}{b-a}t+O\left(t^2\right)$$

I am sure that you can take from here.