1. ## ln Limit Problem

Finally, this looks right:

$\displaystyle \lim x \rightarrow 1$

$\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}$

$\displaystyle \dfrac{5(1)}{(1) - 1} - \dfrac{5}{\ln (1)}$

$\displaystyle \dfrac{5}{0} - \dfrac{5}{0} = \dfrac{0}{0}$ Indeterminate

$\displaystyle \dfrac{5x}{x - 1} - \dfrac{5}{\ln x}$

$\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5(x - 1)}{(x - 1)(\ln x)}$

$\displaystyle \dfrac{5 \ln x}{(x - 1)(\ln x)} - \dfrac{5x - 5}{(x - 1)(\ln x)}$

$\displaystyle \dfrac{5 \ln x - 5x - 5}{\ln x^{2} - \ln x}$

$\displaystyle \dfrac{5 \ln x - 5x - 5}{2 \ln x - \ln x}$

$\displaystyle \dfrac{\dfrac{d}{dx} 5 \ln x - 5x - 5}{\dfrac{d}{dx} 2 \ln x - \ln x}$

$\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{2}{x} - \dfrac{1}{x}}$

$\displaystyle \dfrac{\dfrac{5}{x} - 5}{\dfrac{1}{x}}$

$\displaystyle \dfrac{\dfrac{5}{(1)} - 5}{\dfrac{1}{(1)}} = \dfrac{0}{1} = 0$ Answer?? Computer still says no.

2. ## Re: ln Limit Problem

Hey Jason76.

Hint: You should have [5x*ln(x) - 5(x-1)]/[(x-1)ln(x)] when you simplify both fractions into one compound one. See if this changes your answer.