## Calculus (3rd Edition)

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When $\lim\limits_{x \to \infty} f(x)$ exists then the sequence $a_n=f(n)$ converges to the same limit, so we have: $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} f(x)$ Now, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{\ln (1+\dfrac{1}{x} )^2}{1/\sqrt x}$ We see that this limit shows the indefinite form of $\dfrac{0}{0}$, so we apply L-Hopital's rule. Thus, $\lim\limits_{n \to \infty} a_n =-2 \lim\limits_{x \to \infty} \dfrac{x^{3/2} }{x^2 (1+\dfrac{1}{x})} \\=-2 \lim\limits_{x \to \infty}\dfrac{1}{(1+\dfrac{1}{x})} \cdot \lim\limits_{x \to \infty} \dfrac{1}{\sqrt x}\\= -2 \dfrac{\lim\limits_{x \to \infty} (1)}{\lim\limits_{x \to \infty} (1+\dfrac{1}{x})} \cdot \lim\limits_{x \to \infty} \dfrac{1}{\sqrt x}\\=0$