3.4.238 AP calculus exam Limits with ln

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SUMMARY

The limit expression discussed is $$\displaystyle\lim_{h\to 0} \dfrac{\ln{(4+h)}-\ln{(4)}}{h}$$, which is evaluated using the definition of the derivative. The correct answer is not that the limit diverges, but rather it converges to $$\dfrac{1}{4}$$, as it represents the derivative of the function $$f(x) = \ln(x)$$ at the point $$x=4$$. The initial confusion arose from an incorrect limit expression that included $$\ln(h)$$ instead of $$\ln(4)$$.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with the natural logarithm function
  • Knowledge of the derivative definition
  • Basic graphing skills to visualize functions
NEXT STEPS
  • Study the definition of a derivative in calculus
  • Learn about the properties of logarithmic functions
  • Explore limit evaluation techniques, including L'Hôpital's Rule
  • Practice solving limits involving logarithmic expressions
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Students preparing for AP Calculus exams, educators teaching calculus concepts, and anyone looking to strengthen their understanding of limits and derivatives involving logarithmic functions.

karush
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Solve
$$\displaystyle\lim_{h\to 0}
\dfrac{\ln{(4+h)}-\ln{h}}{h}$$
$$(A)\,0\quad
(B)\, \dfrac{1}{4}\quad
(C)\, 1\quad
(D)\, e\quad
(E)\, DNE$$
The Limit diverges so the Limit Does Not Exist (E)ok the only way I saw that it diverges is by plotting
not sure what the rule is that observation would make ploting not needed

also I noticed these AP sample questions are getting a lot of views so thot I would continue to post more
 
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Check your limit expression again. Should be

$\displaystyle \lim_{h \to 0} \dfrac{\ln(4+h) - \color{red}{ \ln(4)}}{h}$
Recall the limit definition of a derivative ...

$\displaystyle f’(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

let $x=4$

... try again.
 

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