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If $\mathrm{a}>0, \mathrm{~b}>0$ then $\lim _{n \rightarrow \infty}\left(\frac{a+b^{1 / n}-1}{a}\right)^n=$
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$b^{1 / a}$
$y=\lim _{\mathrm{h} \rightarrow \infty}\left|\frac{\mathrm{a}+\mathrm{b}^{\frac{1}{\mathrm{n}}}-1}{\mathrm{a}}\right|$
now, $\log _e|y|=\lim _{\mathrm{h} \rightarrow \infty} n \log _e\left|\frac{a+b^{\frac{1}{2}}-1}{a}\right|^{\mathrm{n}}$
$\Rightarrow \log _{\mathrm{e}}|\mathrm{y}|=\lim _{\mathrm{h} \rightarrow \infty} \mathrm{n}\left[\log _{\mathrm{e}}\left|\mathrm{a}+\mathrm{b}^{\frac{1}{\mathrm{n}}}-1\right|^{\mathrm{n}}-\log _{\mathrm{e}}|\mathrm{a}|\right]$
$\Rightarrow \log _{\mathrm{e}}|y|=\lim _{\mathrm{h} \rightarrow \infty} \mathrm{n}\left[\frac{\log _{\mathrm{e}}\left|\mathrm{a}+\mathrm{b}^{\frac{1}{\mathrm{n}}}-1\right|^{\mathrm{n}}-\log _{\mathrm{e}}|\mathrm{a}|}{(1 / \mathrm{n})}\right]$
$\Rightarrow \log _{\mathrm{e}}|\mathrm{y}|=\lim _{\mathrm{h} \rightarrow \infty} \frac{\left(\frac{1}{\mathrm{a}+\mathrm{b}^{\frac{1}{\mathrm{n}}}-1}\right) \times \mathrm{b}^{\frac{1}{\mathrm{n}}} \times \log _{\mathrm{e}}(\mathrm{b}) \times\left(-\frac{1}{\mathrm{n}^2}\right)}{\left(-\frac{1}{\mathrm{n}^2}\right)}$
$\Rightarrow \log _{\mathrm{e}}|y|=\lim _{\mathrm{b} \rightarrow \infty}\left(\frac{1}{a+b^{\frac{1}{n}}-1}\right) \cdot(b)^{\frac{1}{n}} \times \log _{\mathrm{e}}(b)$
$\Rightarrow \log _{\mathrm{e}}|\mathrm{y}|=\frac{1}{\mathrm{a}} \times \log _{\mathrm{e}}(\mathrm{b})$
$\Rightarrow \log _{\mathrm{e}}|\mathrm{y}|=\log _{\mathrm{e}}(\mathrm{b})^{\frac{1}{2}}$
$\Rightarrow \mathrm{y}(\mathrm{b})^{\frac{1}{\mathrm{a}}}$
now, $\log _e|y|=\lim _{\mathrm{h} \rightarrow \infty} n \log _e\left|\frac{a+b^{\frac{1}{2}}-1}{a}\right|^{\mathrm{n}}$
$\Rightarrow \log _{\mathrm{e}}|\mathrm{y}|=\lim _{\mathrm{h} \rightarrow \infty} \mathrm{n}\left[\log _{\mathrm{e}}\left|\mathrm{a}+\mathrm{b}^{\frac{1}{\mathrm{n}}}-1\right|^{\mathrm{n}}-\log _{\mathrm{e}}|\mathrm{a}|\right]$
$\Rightarrow \log _{\mathrm{e}}|y|=\lim _{\mathrm{h} \rightarrow \infty} \mathrm{n}\left[\frac{\log _{\mathrm{e}}\left|\mathrm{a}+\mathrm{b}^{\frac{1}{\mathrm{n}}}-1\right|^{\mathrm{n}}-\log _{\mathrm{e}}|\mathrm{a}|}{(1 / \mathrm{n})}\right]$
$\Rightarrow \log _{\mathrm{e}}|\mathrm{y}|=\lim _{\mathrm{h} \rightarrow \infty} \frac{\left(\frac{1}{\mathrm{a}+\mathrm{b}^{\frac{1}{\mathrm{n}}}-1}\right) \times \mathrm{b}^{\frac{1}{\mathrm{n}}} \times \log _{\mathrm{e}}(\mathrm{b}) \times\left(-\frac{1}{\mathrm{n}^2}\right)}{\left(-\frac{1}{\mathrm{n}^2}\right)}$
$\Rightarrow \log _{\mathrm{e}}|y|=\lim _{\mathrm{b} \rightarrow \infty}\left(\frac{1}{a+b^{\frac{1}{n}}-1}\right) \cdot(b)^{\frac{1}{n}} \times \log _{\mathrm{e}}(b)$
$\Rightarrow \log _{\mathrm{e}}|\mathrm{y}|=\frac{1}{\mathrm{a}} \times \log _{\mathrm{e}}(\mathrm{b})$
$\Rightarrow \log _{\mathrm{e}}|\mathrm{y}|=\log _{\mathrm{e}}(\mathrm{b})^{\frac{1}{2}}$
$\Rightarrow \mathrm{y}(\mathrm{b})^{\frac{1}{\mathrm{a}}}$
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