Solutions - 0140-01

(1) Indeterminate form:

$$a_n\to \infty \cdot 0\text{as}n\to \infty$$

---

(2) L’Hopital:

Convert:

$$a_n=\sqrt{n}\ln (1+\frac{1}{n})\gg \gg \frac{\ln (1+1/n)}{n^{-1/2}}\to \frac{0}{0}$$

Change to $x$ and apply L’Hopital:

$$\frac{\ln (1+1/x)}{x^{-1/2}}\stackrel{d/dx}{⟶}\frac{\frac{-1/x^2}{1+1/x}}{-1/2\cdot x^{-3/2}}\gg \gg \frac{2}{x^{+1/2}+x^{-1/2}}$$

---

(3) Take limit:

$$\frac{2}{x^{-1/2}+x^{+1/2}}⟶0\text{as}x\to \infty$$

Therefore $a_n\to 0$ as $n\to \infty$.