How can i prove this interesting limit Was just playing around any thought it was interesting.
knowing now that $a>b>0$
$\lim\limits_{x\rightarrow \infty} (a^x-b^x)^{\frac{a+b}{x}} = a^{a+b}$
 A: We have
$$(a^x - b^x)^{(a + b)/x} = b^{a+b}\left(\left(\frac{a}{b}\right)^x - 1\right)^{(a + b)/x}.$$
For any real $t > 1$, we have $(t^x - 1)^{1/x} \to t$ as $x\to\infty$, since
$$\log\left[(t^x - 1)^{1/x}\right] = \frac{1}{x} \log (e^{x \log t}- 1) \to \log t$$
as $x \to \infty$ (from L'Hopital's rule, a power series expansion, treating the approximation $\log(e^{x \log t} - 1) \approx (\log e^{x \log t}) = x\log t$ carefully, etc.). Thus as $x \to \infty$, we have
$$b^{a+b}\left(\left(\frac{a}{b}\right)^x - 1\right)^{(a + b)/x} \to b^{a + b}\left(\frac{a}{b}\right)^{a + b} = a^{a+b}.$$
A: Just using the relationship between $\exp$ and $\ln$, and the relationship between limits and continuous functions:
$$\begin{align}
\lim_{x\to\infty}\left(a^x-b^x\right)^{\frac{a+b}{x}}&=\left(e^{\lim\limits_{x\to\infty}\frac1{x}\ln\left(a^x-b^x\right)}\right)^{a+b}
\end{align}$$
Can you show the limit in the exponent is $\ln(a)$? L'Hospital's rule will work.
$$
\begin{align}
\lim\limits_{x\to\infty}\frac1{x}\ln\left(a^x-b^x\right)
&\stackrel{LH}{=}\lim\limits_{x\to\infty}\frac{a^x\ln(a)-b^x\ln(b)}{a^x-b^x}\\
&=\lim\limits_{x\to\infty}\frac{(a/b)^x\ln(a)-\ln(b)}{(a/b)^x-1}\\
&\stackrel{LH}{=}\lim\limits_{x\to\infty}\frac{(a/b)^x\ln(a)\ln(a/b)}{(a/b)^x\ln(a/b)}\\
&=\ln(a)
\end{align}$$
A: we show: $g(x) = \left(1-\left(\dfrac{b}{a}\right)^x\right)^{\dfrac{a+b}{x}} \longrightarrow 1$ or $\dfrac{a+b}{x}\cdot \ln\left(1-\left(\dfrac{b}{a}\right)^x\right) \longrightarrow 0$. To this end, we can use L'hopital rule to show that: $f(x) =\dfrac{\ln\left(1-\left(\dfrac{b}{a}\right)^x\right)}{x} \longrightarrow 0$. Apply L'hopital rule we have:
$f(x) \longrightarrow \displaystyle \lim_{x \to \infty}\dfrac{-\ln\left(\dfrac{b}{a}\right)\cdot \left(\dfrac{b}{a}\right)^x}{1-\left(\dfrac{b}{a}\right)^x} = 0$. So $\displaystyle \lim_{x \to \infty} LHS = a^{a+b}\cdot \displaystyle \lim_{x \to \infty} g(x) = a^{a+b}$
A: It can be done as follows:
$$
\lim_{x\to\infty} \left(a^x-b^x\right)^{\frac{a+b}{x}}=\lim_{x\to\infty} a^{a+b}\cdot\left(1-\left(\frac{b}{a}\right)^x\right)^{\frac{a+b}{x}}=
$$
Now we have $\lim_{x\to\infty} \left(\frac{b}{a}\right)^x=0$ and $\lim_{x\to\infty} \frac{a+b}{x}=0$ and therefore:
$$
\lim_{x\to\infty} a^{a+b}\cdot\left(1-\left(\frac{b}{a}\right)^x\right)^{\frac{a+b}{x}}=a^{a+b}\cdot\left(1-0\right)^0=a^{a+b}
$$
