Question:
If \( f(3) = 18, f'(3) = 0 \) and \( f''(3) = 4 \), then the value of
\[
\lim_{x \to 3} \ln \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^3}}
\]
is equal to
If \( f(3) = 18, f'(3) = 0 \) and \( f''(3) = 4 \), then the value of
\[
\lim_{x \to 3} \ln \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^3}}
\]
is equal to
Show Hint
For limits involving logarithms, simplify the expression using logarithmic properties and apply L'Hopital's Rule for indeterminate forms.
Updated On: Jan 23, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Use the given limit.
We are given the limit: \[ \lim_{x \to 3} \ln \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^3}} \] We can simplify the expression using the properties of logarithms: \[ \ln \left( \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^3}} \right) = \frac{18}{(x-3)^3} \ln \left( \frac{f(x+2)}{f(3)} \right) \] Step 2: Apply L'Hopital's Rule.
Since this is an indeterminate form of type \( \frac{0}{0} \), we can apply L'Hopital's Rule. Differentiating the numerator and denominator gives the final answer as \( 2 \).
We are given the limit: \[ \lim_{x \to 3} \ln \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^3}} \] We can simplify the expression using the properties of logarithms: \[ \ln \left( \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^3}} \right) = \frac{18}{(x-3)^3} \ln \left( \frac{f(x+2)}{f(3)} \right) \] Step 2: Apply L'Hopital's Rule.
Since this is an indeterminate form of type \( \frac{0}{0} \), we can apply L'Hopital's Rule. Differentiating the numerator and denominator gives the final answer as \( 2 \).
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