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anonymous
 4 years ago
limit of x> 1+
{ x/(x1)  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 = ...
anonymous
 4 years ago
limit of x> 1+ { x/(x1)  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 = ...

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 1+}\left( (x/x1) (1/lnx)\right)\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow 1^+} \frac{\ln(x) \cdot x(x1)}{\ln(x) \cdot (x1)}\] \[\lim_{x \rightarrow 1^+} \frac{\frac{1}{x} \cdot x+\ln(x) \cdot 1(10)}{\frac{1}{x}(x1)+\ln(x)(10)}\] 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! :)

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{\frac{1}{1}}{\frac{1}{1^2}+\frac{1}{1}}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{1+1}=\frac{1}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm, seems correct, i'll try and caculate again, thanks :) i'll let you know if somethings up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is correct :D made a few mistakes with the chain rule, learned a lot from you
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