## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}\Bigg(1+\frac{a}{x}\Bigg)^{bx}=e^{ab}$$
$$A=\lim_{x\to\infty}\Bigg(1+\frac{a}{x}\Bigg)^{bx}$$ We take the natural logarithm of both sides: $$\ln A=\ln\Bigg[\lim_{x\to\infty}\Bigg(1+\frac{a}{x}\Bigg)^{bx}\Bigg]$$ $$\ln A=\lim_{x\to\infty}\Bigg[\ln\Bigg(1+\frac{a}{x}\Bigg)^{bx}\Bigg]$$ $$\ln A=\lim_{x\to\infty}bx\ln \Bigg(1+\frac{a}{x}\Bigg)$$ $$\ln A=\lim_{x\to\infty}\frac{bx\ln\Bigg(1+\frac{a}{x}\Bigg)}{1}.$$ Divide both numerator and denominatory by $x$ $$\ln A=\lim_{x\to\infty}\frac{b\ln(1+\frac{a}{x})}{\frac{1}{x}}.$$ Now we take $\frac{1}{x}=u$. So as $x\to\infty$, we have $u\to0$. $$\ln A=\lim_{u\to0}\frac{b\ln(1+au)}{u}$$ Because $\lim_{u\to0}[b\ln(1+au)]=b\ln(1+a\times0)=b\ln1=b\times0=0$ and $\lim_{u\to0}u=0$, we have an indeterminate form of $\frac{0}{0}$. Following L'Hospital's Rule, we would have $$\ln A=\lim_{u\to0}\frac{b\times\frac{1}{1+au}\times(1+au)'}{1}$$ $$\ln A=\lim_{u\to0}\frac{b}{1+au}\times a$$ $$\ln A=\lim_{u\to0}\frac{ab}{1+au}$$ $$\ln A=\frac{ab}{1+a\times0}$$ $$\ln A=\frac{ab}{1}=ab$$ $$A=e^{ab}$$