Question-170321

Question Number 170321 by amin96 last updated on 20/May/22

Answered by aleks041103 last updated on 21/May/22

a)

\frac{2^{x}}{3^{x^{2}}} = 2^{x} 3^{- x^{2}} = e^{l n (2) x - l n (3) x^{2}}

\underset{x \to \infty}{l i m} \frac{2^{x}}{3^{x^{2}}} = {\underset{x \to \infty}{l i m e}}^{l n (2) x - l n (3) x^{2}} = e^{- \infty} = 0

b)

\underset{x \to 0^{+}}{l i m} \frac{e^{- 3 / x}}{x^{2}} = [\frac{0}{0}] = \underset{1 / x \to \infty}{l i m} {(\frac{1}{x})}^{2} e^{- 3 (1 / x)}

= \underset{y \to \infty}{l i m} \frac{y^{2}}{e^{3 y}} = \underset{y \to \infty}{l i m} \frac{2 y}{3 e^{3 y}} = \underset{y \to \infty}{l i m} \frac{2}{9 e^{3 y}} = 0

c)

\underset{x \to \infty}{l i m} \frac{{l n}^{5} x}{x^{2}} = {(\underset{x \to \infty}{l i m} \frac{l n x}{x^{2 / 5}})}^{5} = {(\underset{x \to \infty}{l i m} \frac{l n x}{x^{2 / 5}})}^{5} =

= {(\frac{5}{2} \underset{x \to \infty}{l i m} \frac{l n (x^{2 / 5})}{x^{2 / 5}})}^{5} = \frac{5^{5}}{2^{5}} {(\underset{y \to \infty}{l i m} \frac{l n y}{y})}^{5} =

= \frac{625 \times 5}{32} {(\underset{y \to \infty}{l i m} \frac{1 / y}{1})}^{5} = 0

d)

b y a n a l o g y \to 0