Question

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$\lim_{x \to \infty}(\frac{n+x}{n-1})^n$

Answer 1

$\displaystyle \lim_{x \to \infty} \bigg(\frac{n+x}{n-1}\bigg)^n=\left\{\begin{array}{ccc}n+x=t\\ t\to \infty\\ n=t-x\end{array}\right\}=\lim_{t \to \infty}\bigg(\frac{t}{t-x-1}\bigg)^{t-x}=\{1^{\infty}\}=\\ \\ \\ =\lim_{t \to \infty}\bigg(1+\frac{x+1}{t-x-1}\bigg)^\big{(t-x\bigg)\cdot\frac{x+1}{t-x-1}\cdot\frac{t-x-1}{x+1}}=\\ \\ \\ =e^\bigg{\displaystyle \lim_{t \to \infty}\frac{(t-x)(x+1)}{t-x-1}}=e^\bigg{\displaystyle\lim_{t \to \infty}\frac{(1-\frac{x}{t})(x+1)}{1-\frac{x}{t}-\frac{1}{t}}}=e^\bigg{\displaystyle\lim_{t \to \infty}\frac{(1-0)\cdot(x+1)}{1-0-0}}=e^{x+1}$