Find limit $1^\infty$

Set the limit equal to $y$

Taking $\text{ln()}$ both side we have

$\displaystyle \lim_{n \to \infty} n \ln \left(\dfrac{a+b^{1/n}-1}{a}\right) = \ln (y)$

$\displaystyle \lim_{n \to \infty}\dfrac{\ln \left(\dfrac{a+b^{1/n}-1}{a}\right)}{1/n} = \ln (y)$

Which converts the limit to the $\color{#D61F06}{\dfrac{0}{0}}$ form,now I will be using L Hospital rule

$\displaystyle \lim_{n \to \infty} \dfrac{\left(\dfrac{a}{a+b^{1/n}-1}\right)\dfrac{-b^{1/n}\ln (b)}{a\times{n^2}}}{\left(-\dfrac{1}{n^2}\right)} = \ln (y)$

On simplifying we get

$\displaystyle \lim_{n \to \infty} \dfrac{b^{1/n} \ln (b)}{a+b^{1/n}-1}=\dfrac{\ln (b)}{a}=\ln b^{1/a}=\ln (y)$

So $y=e^{\ln (b^{1/a})}=\color{#3D99F6}{\boxed{\sqrt[a] b}}$

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I have used the tool that differentiation of $b^{1/n}$ is $\dfrac{-b^{1/n}\ln (b)}{n^2}$

$\color{#D61F06}{\text{METHOD 2}}$

Assume $b=0$ then limit must equal to $0$ as $\alpha^{\infty}=0$ when $\alpha \in [0,1)$

and only one option gives 0 when we put b=0 that is our answer.

Let, $y=\lim_{n \to \infty} \left(\dfrac{a+\sqrt[n]{b}-1}{a}\right)^{n}$ taking $\ln_{}{}$ of both sides, $\ln{y}=\lim_{n \to \infty}{n\times \ln{\left(\frac{a+\sqrt[n]{b}-1}{a}\right)}}$ Substitute $t=\frac{1}{n}$ $\ln{y}=\lim_{t \to 0}{\frac{\left(\ln{\frac{a+b^{t}-1}{a}}\right)}{t}}$ Since, this limit is in $\frac{0}{0}$ form, we may use L'Hôpital's rule $\large \ln{y}=\lim_{t \to 0}\frac{\frac{a}{a+b^{t}-1}\times \frac{\ln{b}\times b^{t}}{a}}{1}=\frac{\ln{b}}{a}$ $\Rightarrow y=\sqrt[a]{b}$

You can use expansion of $b^{1/n}$

$b^{1/n}$ = $1+\frac{ln(b)}{n}$ and then binomial expansion