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.