limit : $\lim_{n\to \infty} (1+\frac{\sqrt[n]{a}-1}{b})^n=?$ fine the limit :
$$\lim_{n\to \infty} (1+\frac{\sqrt[n]{a}-1}{b})^n=?$$
$a,b\in \mathbb{R}$
My Try :
$$\lim_{n\to \infty} (1)+\lim_{n\to \infty}(\frac{\sqrt[n]{a}-1}{b})^n=1+0=1$$
Because:
$$\lim_{n\to \infty} \sqrt[n]{a}=1 $$
is it right ?
 A: No, this is not right.


*

*Your first step:
$$
\lim_{n\to \infty} (1+\frac{\sqrt[n]{a}-1}{b})^n
= \lim_{n\to \infty} 1
+ \lim_{n\to \infty} (\frac{\sqrt[n]{a}-1}{b})^n 
$$
is wrong. With the same argument, you would have $$\lim_{n\to\infty} (1+1)^n = \lim_{n\to\infty} 1^n + \lim_{n\to\infty} 1^n = 1+1 = 2$$ which is clearly false.

*Now, let us prove that the limit is $a^{1/b}$. For everything to make sense to begin with, one must have $a,b>0$. I will only use elementary arguments involving the fact that $\frac{d}{dx}e^x\mid_{x=0} = 1$ and $\frac{d}{dx}\ln(1+x)\mid_{x=0} = 1$.
To begin with, let us put our quantity is the simpler to analyze exponential form: 
$$
(1+\frac{\sqrt[n]{a}-1}{b})^n
= e^{n \ln\left(1+\frac{\sqrt[n]{a}-1}{b}\right)} 
= \exp\left({n \ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}\right)
$$
so by continuity of $\exp$, it suffices to find the the limit of ${n \ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}$.
Since $\frac{e^{\frac{\ln a}{n}}-1}{b}\xrightarrow[n\to\infty]{0}$ and $\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1$, we can rewrite
$$
{n \ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}
= n \cdot \frac{e^{\frac{\ln a}{n}}-1}{b} \cdot \underbrace{ \frac{\ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}{\frac{e^{\frac{\ln a}{n}}-1}{b}}}_{\to_{n\to\infty} 1}
$$
and thus focus on the limit of $n \cdot \frac{e^{\frac{\ln a}{n}}-1}{b}$.
Recalling that $1=\exp'(0) = \lim_{x\to 0} \frac{e^x-1}{x}$ and that $\frac{\ln a}{n}\xrightarrow[n\to\infty]{} 0$, we then can (!) write
$$n \cdot \frac{e^{\frac{\ln a}{n}}-1}{b}
=n\cdot\frac{\frac{\ln a}{n}}{b} \cdot \frac{e^{\frac{\ln a}{n}}-1}{\frac{\ln a}{n}}
= \frac{\ln a}{b} \cdot \frac{e^{\frac{\ln a}{n}}-1}{\frac{\ln a}{n}}
\xrightarrow[n\to\infty]{}\frac{\ln a}{b}\cdot 1.
$$
Putting it all together,
$$
(1+\frac{\sqrt[n]{a}-1}{b})^n
= \exp\left({n \ln\left(1+\frac{e^{\frac{\ln a}{n}}-1}{b}\right)}\right)
\xrightarrow[n\to\infty]{} 
\exp\left(\frac{\ln a}{b}\cdot 1\cdot 1\right)
= e^{\frac{\ln a}{b}}
$$
giving that the limit is $e^{\frac{\ln a}{b}}=\boxed{a^{\frac{1}{b}}}$.
A: Let $L$ be the limit.  Then,
$$\ln(L)=\lim_{n\to\infty}\frac{\ln\left(1+\frac{\sqrt[n]a-1}b\right)}{1/n}\stackrel{L'H}=\lim_{n\to\infty}\frac{(-1/n^2)\frac{\ln(a)}b\sqrt[n]a}{\left(1+\frac{\sqrt[n]a-1}b\right)(-1/n^2)}\\=\frac{\ln(a)}b\lim_{n\to\infty}\frac1{\left(\frac1{\sqrt[n]a}+\frac{1-\frac1{\sqrt[n]a}}b\right)}\\=\frac{\ln(a)}b$$
Thus,
$$L=e^{\ln(a)/b}=\sqrt[b]a$$
