Calculating the limit $\lim_{x \to \infty } \left( \frac{a-1+b ^{ \frac{1}{x} } }{a} \right) ^{x}$ for $a>0, b>0$ If $a>0, b>0$ then what is the limit
$$\lim_{x \to \infty  } \left(  \frac{a-1+b ^{ \frac{1}{x} } }{a} \right)  ^{x}$$
I tried putting $y=\frac{1}{x}$ but it's not working.
 A: The idea of setting $x=1/y$ is good; go ahead by taking the logarithm:
$$
\log\left(\frac{a-1+b^y}{a}\right)^{\!1/y}=\frac{\log(a-1+b^y)-\log a}{y}
$$
so you want to compute
$$
\lim_{y\to0^+}\frac{\log(a-1+b^y)-\log a}{y}
$$
This is the derivative at $0$ of $f(y)=\log(a-1+b^y)$; since
$$
f'(y)=\frac{b^y\log b}{a-1+b^y}
$$
we have
$$
f'(0)=\frac{\log b}{a}=\log(b^{1/a})
$$
Therefore your limit is
$$
\exp(\log(b^{1/a})=b^{1/a}
$$
A: Best way is to take log and proceed
(1+ (b^(1/x)-1)/a)^x=y
lny=xln(1+(b^(1/x)-1)/a))
open the log taylor series (only 1 term would suffice) (with (b^(1/x)-1)/a tending zero)
lny=x((b^(1/x)-1)/a)
open exponential taylor series (again one term would suffice) (with 1/x tending zero)
lny=x(1/x)(lnb)/a
lny=lnb/(a)
y=b^(1/a)=Limit needed
OR
lny=ln(1+(b^(1/x)-1)/a)/(1/x)
lny={(-1/x^2) (lnb) [(b^1/x)/a] / [1+(b^(1/x)-1)/a]}/[-1/x^2]
lny=lnb/a
A: Some well known facts:
1) $e^u = 1+u+o(u)$ as $u\to 0.$ This follows just from the fact that $\exp'(0) = 1.$
2) $\lim_{x\to\infty}(1+c/x + o(1/x))^x = e^c.$ Here $c$ is a constant. (This can be derived by applying $\ln,$ and using $\ln '(1) = 1.$)
From 1) we get $b^{1/x} = e^{(\ln b)/x} = 1 + (\ln b)/x + o((\ln b)/x) = 1 + (\ln b)/x + o(1/x).$ Thus our expression equals
$$(1 +[(\ln b)/a]/x + o(1/x))^x,$$
which by 2) has limit $e^{(\ln b)/a} = b^{1/a}.$
