redupticslaz

2022-04-25

How can I show that $\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{log}(x+1)}{\mathrm{log}x}=1?$

bailaretzy33

Beginner2022-04-26Added 15 answers

You could use l'Hospital's rule:

$\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{log}(x+1)}{\mathrm{log}\left(x\right)}=\underset{x\to \mathrm{\infty}}{lim}\frac{\frac{1}{x+1}}{\frac{1}{x}}=\underset{x\to \mathrm{\infty}}{lim}\frac{1}{\frac{x+1}{x}}=\underset{x\to \mathrm{\infty}}{lim}\frac{1}{1+\frac{1}{x}}$

$=\frac{1}{1+\underset{x\to \mathrm{\infty}}{lim}\frac{1}{x}}=\frac{1}{1+0}=1$

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