Types of Infinity — Question 10

Question 10

Prove that $$\lim_{x\to\infty}\left(\frac{x}{x+1}\right)^x=\frac{1}{e}.$$

Question 10 - Solution

Let $$L=\lim_{x\to\infty}\left(\frac{x}{x+1}\right)^x.$$

Take the natural logarithm: $$\ln L=\lim_{x\to\infty} x\ln\!\left(\frac{x}{x+1}\right).$$

Rewrite the logarithm: $$\ln\!\left(\frac{x}{x+1}\right) = \ln\!\left(\frac{1}{1+\frac{1}{x}}\right) = -\ln\!\left(1+\frac{1}{x}\right).$$

Thus, $$\ln L = -\lim_{x\to\infty} x\ln\!\left(1+\frac{1}{x}\right).$$

Consider the limit $$\lim_{x\to\infty} x\ln\!\left(1+\frac{1}{x}\right).$$

Rewrite it as a quotient: $$\lim_{x\to\infty} \frac{\ln\!\left(1+\frac{1}{x}\right)}{\frac{1}{x}}.$$

This is an indeterminate form $\frac{0}{0}$, so apply L’Hôpital’s Rule.

Differentiate numerator and denominator: $$\frac{d}{dx}\ln\!\left(1+\frac{1}{x}\right) = \frac{-1}{x(x+1)}, \qquad \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}.$$

Thus, $$\lim_{x\to\infty} \frac{\ln\!\left(1+\frac{1}{x}\right)}{\frac{1}{x}} = \lim_{x\to\infty} \frac{\frac{-1}{x(x+1)}}{\frac{-1}{x^2}} = \lim_{x\to\infty}\frac{x^2}{x(x+1)} = \lim_{x\to\infty}\frac{x}{x+1} = 1.$$

Therefore, $$\ln L=-1.$$

Exponentiating both sides: $$L=e^{-1}=\frac{1}{e}.$$

$$\boxed{\frac{1}{e}}$$