Solutions - 0140-07

(a) $\ln (2/3)$

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(b) diverges:

$$\frac{e^n}{2^n}\gg \gg {\left(\frac{e}{2}\right)}^n\stackrel{n\to \infty }{⟶}\infty$$

(Note that $e/2>1$.)

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(c) $0$:

L’Hopital’s Rule:

$$\begin{array}{c}\gg \gg \frac{{(\ln x)}^2}{x}{\underset{d/dx}{⟶}}^{d/dx}\frac{\frac{2}{x}\ln x}{1}=\frac{2\ln x}{x}\\ \\ {\underset{d/dx}{⟶}}^{d/dx}\frac{2/x}{1}=\frac{2}{x}\stackrel{n\to \infty }{⟶}0\end{array}$$

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(d) $0$ by (c), the sign doesn’t affect convergence to $0$

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(e) $-{\displaystyle \frac{1}{7}}$:

Multiply above and below by $4^{-n}$:

$$\begin{array}{c}\frac{3-4^n}{2+7\cdot 4^n}\gg \gg \frac{4^{-n}}{4^{-n}}\cdot \frac{3-4^n}{2+7\cdot 4^n}\\ \\ \gg \gg \frac{-1+\frac{3}{4^n}}{7+\frac{2}{4^n}}\stackrel{n\to \infty }{⟶}-\frac{1}{7}\end{array}$$

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(f) $e$:

This is a well-known formula for $e$. If that formula is not used as the definition of $e$, then it would not be circular reasoning to argue as follows:

$$\begin{array}{c}\underset{n\to \infty }{\lim }{(1+\frac{1}{n})}^n\gg \gg \underset{x\to \infty }{\lim }\exp \left(\ln {(1+\frac{1}{x})}^x\right)\\ \\ \gg \gg \exp (\underset{x\to \infty }{\lim }\ln {(1+\frac{1}{x})}^x)\gg \gg \exp (\underset{x\to \infty }{\lim }\frac{\ln (1+\frac{1}{x})}{1/x})\\ \\ \gg \gg \exp (\underset{x\to \infty }{\lim }\frac{-\frac{1}{x^2}\cdot \frac{1}{1+1/x}}{-1/x^2})\gg \gg \exp (\underset{x\to \infty }{\lim }\frac{-\frac{1}{x^2}\cdot \frac{1}{1+1/x}}{-1/x^2})\\ \\ \gg \gg \exp (1)\end{array}$$

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(g) diverges:

$$\ln (1+\frac{1}{n})\stackrel{n\to \infty }{⟶}\ln (1^{+})=0^{+}$$

So:

$$\frac{1}{\ln (1+\frac{1}{n})}\stackrel{n\to \infty }{⟶}\frac{1}{0^{+}}=+\infty$$