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anonymous

  • 4 years ago

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 = ...

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  1. anonymous
    • 4 years ago
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    \[\lim_{x \rightarrow 1+}\left( (x/x-1) -(1/lnx)\right)\]

  2. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 1^+} \frac{\ln(x) \cdot x-(x-1)}{\ln(x) \cdot (x-1)}\] \[\lim_{x \rightarrow 1^+} \frac{\frac{1}{x} \cdot x+\ln(x) \cdot 1-(1-0)}{\frac{1}{x}(x-1)+\ln(x)(1-0)}\] 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! :)

  3. myininaya
    • 4 years ago
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    \[\frac{\frac{1}{1}}{\frac{1}{1^2}+\frac{1}{1}}\]

  4. myininaya
    • 4 years ago
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    \[\frac{1}{1+1}=\frac{1}{2}\]

  5. anonymous
    • 4 years ago
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    hmm, seems correct, i'll try and caculate again, thanks :) i'll let you know if somethings up

  6. anonymous
    • 4 years ago
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    it is correct :D made a few mistakes with the chain rule, learned a lot from you

  7. myininaya
    • 4 years ago
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    :)

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