Need help evaulating a few limits

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SUMMARY

This discussion focuses on evaluating limits in the context of a chemistry problem involving the pH of a solution as a function of the volume of titrant added. The limits presented include \(\lim_{V→50^{-}}-log(\frac{50-V}{10(V+50)})\) and \(\lim_{V→50^{+}}log(\frac{V-50}{10(V+50)})\). The analysis concludes that L'Hopital's Rule is not applicable in this scenario, as the denominator does not approach zero. Instead, the limit as \(V\) approaches 50 from the left results in \(-\infty\).

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  • Understanding of limits in calculus
  • Familiarity with logarithmic functions
  • Knowledge of L'Hopital's Rule
  • Basic principles of pH and titration in chemistry
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  • Study the application of L'Hopital's Rule in various limit problems
  • Explore piecewise continuous functions in calculus
  • Investigate the relationship between pH and titrant volume in acid-base titrations
  • Learn about continuity and limits of logarithmic functions
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Chemistry students, calculus learners, and anyone interested in understanding limits and their applications in chemical titration scenarios.

Bipolarity
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I made up this limit problems trying to understand something in chemistry, so they may not necessarily have any answer.

I am not sure how to evaluate them analytically. Not sure how L'Hopital's Rule would work here either.

[tex](\lim_{V→50^{-}}-log(\frac{50-V}{10(V+50)}))[/tex]

[tex]14+(\lim_{V→50^{+}}log(\frac{V-50}{10(V+50)}))[/tex]

[tex]14+(\lim_{V→∞}log(\frac{V-50}{10(V+50)}))[/tex]

I would greatly appreciate all your help on these questions!

Just in case anyone asks, I am trying to model the pH of a solution as a function of the volume of titrant added, but I want to make sure it is piece-wise continuous.

Thanks!

BiP
 
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log(x) is continuous for x> 0 so [itex]\lim_{V\to 50^-} log((50- V)/(10(V+ 50))= log(\lim_{V\to 50^-} (50-V)/(10(50+ V))[/itex]

And since the denominator does not go to 0 as V goes to 50, you cannot use "L'Hopital's rule" nor do you need to. As V goes to 50, from below, the numerator goes to 0 while the denominator goes to 1000 so the argument of log goes to 0 and so the limit is [itex]-\infty[/itex].
 

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