Limit Problem - # 5

[Jason76]

\(\displaystyle \lim x \rightarrow \infty\)

\(\displaystyle \dfrac{(\ln x)^{2}}{5x}\)

\(\displaystyle \dfrac{(\ln (\infty))^{2}}{5(\infty)} = \dfrac{\infty}{\infty}\) - indeterminate

\(\displaystyle \dfrac{(\ln x)^{2}}{5x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} (\ln x)^{2}}{\dfrac{d}{dx} 5x}\)

\(\displaystyle \dfrac{2 u}{5}\)

\(\displaystyle \dfrac{2 (\dfrac{1}{x})}{5}\)

\(\displaystyle \dfrac{ (\dfrac{2}{x})}{5}\)

\(\displaystyle \dfrac{ (\dfrac{2}{\infty})}{5} = \dfrac{0}{5} = 0\) answer?

 

[Jason76]

Some mistake with substitution:

\(\displaystyle \lim x \rightarrow \infty\)

\(\displaystyle \dfrac{(\ln x)^{2}}{5x}\)

\(\displaystyle \dfrac{(\ln (\infty))^{2}}{5(\infty)} = \dfrac{\infty}{\infty}\) - indeterminate

\(\displaystyle \dfrac{(\ln x)^{2}}{5x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} (\ln x)^{2}}{\dfrac{d}{dx} 5x}\)

\(\displaystyle \dfrac{2 u}{5}\)

\(\displaystyle \dfrac{2 [u \dfrac{1}{v} (1)]}{5}\)

\(\displaystyle \dfrac{2 [u \dfrac{1}{v} ]}{5}\)

\(\displaystyle \dfrac{2 u \dfrac{1}{x}]}{5}\)

\(\displaystyle \dfrac{u \dfrac{2}{x})}{5}\)

\(\displaystyle \dfrac{\ln x(\dfrac{2}{x})}{5}\)

\(\displaystyle \dfrac{\ln (\infty)(\dfrac{2}{\infty})}{5}\)

\(\displaystyle \dfrac{(\infty) (0)}{5}\)

 

[HallsofIvy]

Some mistake with substitution:

\(\displaystyle \lim x \rightarrow \infty\)

\(\displaystyle \dfrac{(\ln x)^{2}}{5x}\)

\(\displaystyle \dfrac{(\ln (\infty))^{2}}{5(\infty)} = \dfrac{\infty}{\infty}\) - indeterminate

\(\displaystyle \dfrac{(\ln x)^{2}}{5x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} (\ln x)^{2}}{\dfrac{d}{dx} 5x}\)

\(\displaystyle \dfrac{2 u}{5}\)

This is meaningless because you haven't said what "u" is. If you intended "u= ln(x)" so \(\displaystyle (ln(x))^2= u^2\) then \(\displaystyle \dfrac{d (ln(x))^2}{dx}= \dfrac{u^2}{du}\dfrac{du}{dx}\) which is NOT "2u".

\(\displaystyle \dfrac{2 [u \dfrac{1}{v} (1)]}{5}\)

\(\displaystyle \dfrac{2 [u \dfrac{1}{v} ]}{5}\)

\(\displaystyle \dfrac{2 u \dfrac{1}{x}]}{5}\)

\(\displaystyle \dfrac{u \dfrac{2}{x})}{5}\)

\(\displaystyle \dfrac{\ln x(\dfrac{2}{x})}{5}\)

\(\displaystyle \dfrac{\ln (\infty)(\dfrac{2}{\infty})}{5}\)

\(\displaystyle \dfrac{(\infty) (0)}{5}\)