Types of Infinity — Question 8

Question 8 - Solution

Combine the logarithms using logarithmic properties: $$\ln(x+1)-\ln x=\ln\!\left(\frac{x+1}{x}\right).$$

Rewrite the fraction: $$\frac{x+1}{x}=1+\frac{1}{x}.$$

Thus, the expression becomes $$\ln\!\left(1+\frac{1}{x}\right).$$

Now take the limit as $x\to\infty$. Since $$\lim_{x\to\infty}\frac{1}{x}=0,$$ and the natural logarithm function is continuous at $1$, we obtain $$\lim_{x\to\infty}\ln\!\left(1+\frac{1}{x}\right)=\ln(1)=0.$$

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