Question

limit of x-> 1+
{ x/(x-1) - 1/lnx} Anyone know how to solve this?
what i did is making the denominator the same so i can use the L'hopital rule, but then when i use it i get 1/0 = ...

Answer 1

\[\lim_{x \rightarrow 1+}\left( (x/x-1) -(1/lnx)\right)\]

Answer 2

\[\lim_{x \rightarrow 1^+} \frac{\ln(x) \cdot x-(x-1)}{\ln(x) \cdot (x-1)}\]
\[\lim_{x \rightarrow 1^+} \frac{\frac{1}{x} \cdot x+\ln(x) \cdot 1-(1-0)}{\frac{1}{x}(x-1)+\ln(x)(1-0)}\]
i applied l'hospital there
\[\lim_{x \rightarrow 1^+}\frac{1+\ln(x)-1}{1-\frac{1}{x}+\ln(x)}\]
\[\lim_{x \rightarrow 1^+}\frac{\ln(x)}{1-\frac{1}{x}+\ln(x)}\]
we still have 0/0 so we can still use l'hospital
\[\lim_{x \rightarrow 1^+}\frac{\frac{1}{x}}{0-\frac{-1}{x^2}+\frac{1}{x}}\]
\[\lim_{x \rightarrow 1^+}\frac{\frac{1}{x}}{\frac{1}{x^2}+\frac{1}{x}} \]
Now you can plug in 1! :)

Answer 3

\[\frac{\frac{1}{1}}{\frac{1}{1^2}+\frac{1}{1}}\]

Answer 4

\[\frac{1}{1+1}=\frac{1}{2}\]

Answer 5

hmm, seems correct, i'll try and caculate again, thanks :) i'll let you know if somethings up

Answer 6

it is correct :D made a few mistakes with the chain rule, learned a lot from you

Answer 7

:)