# Recitation 13

Exercise: Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

1. $\lim_{t\to 1}\frac{t^8-1}{t^5-1}$
2. $\lim_{x\to 0}\frac{x3^x}{3^x-1}$
3. $\lim_{x\to 0}\cot 2x\sin 6x$
4. $\lim_{x\to \infty}(x-\ln x)$
1. By l’Hospital’s Rule, $\lim_{t\to 1}\frac{t^8-1}{t^5-1}=\lim_{t\to 1}\frac{8t^7}{5t^4}=\frac{8}{5}$.
2. By l’Hospital’s Rule, $\lim_{x\to 0}\frac{x3^x}{3^x-1}=\lim_{x\to 0}\frac{3^x+x3^x\ln3}{3^x\ln3}=\lim_{x\to 0}\frac{1}{\ln3}+x=\frac{1}{\ln3}$.
3. Since $\lim_{x\to 0}\cot 2x\sin 6x=\lim_{x\to 0}\frac{\sin 6x}{\tan 2x}$, by l’Hospital’s Rule, $\lim_{x\to 0}\frac{\sin 6x}{\tan 2x}=\lim_{x\to 0}\frac{6\cos 6x}{2\sec^2 2x}=3$
4. Since $x-\ln x=x(1-\frac{\ln x}{x})$ and by l’Hospital’s Rule $\lim_{x\to \infty}\frac{\ln x}{x}=\lim_{x\to \infty}\frac{1/x}{1}=0$, $\lim_{x\to \infty}(x-\ln x)=\lim_{x\to \infty}x(1-\frac{\ln x}{x})=\lim_{x\to \infty}x=\infty$

Exercise: Use a graph to estimate the value of the limit. Then use l’Hospital’s Rule to find the exact value. $\lim_{x\to\infty}(1+\frac{2}{x})^x$.

Since
$$(1+\frac{2}{x})^x=e^{x\ln(1+\frac{2}{x})}$$
and by l’Hospital’s Rule
$$\lim_{x\to\infty}x\ln(1+\frac{2}{x})=\lim_{x\to\infty}\frac{\ln(1+\frac{2}{x})}{\frac{1}{x}}=\lim_{x\to\infty}\frac{\frac{1}{1+\frac{2}{x}}\frac{-2}{x^2}}{-\frac{1}{x^2}}=2,$$
$$\lim_{x\to\infty}(1+\frac{2}{x})^x=e^2.$$