91Ó°ÊÓ

When encountering limits, particularly those that result in indeterminate forms such as \[ \frac{0}{0} \text{ or } \frac{\text{∞}}{\text{∞}} \], L'Hôpital's Rule becomes a valuable tool. This rule allows us to find limits by differentiating the numerator and the denominator independently. Let’s break down how you can apply L'Hôpital's Rule: L'Hôpital's Rule states that if \[ \text{lim}_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{0}{0} \text{ or } \frac{\text{∞}}{\text{∞}}, \] then you can evaluate this limit by:\[ \text{lim}_{x \rightarrow c} \frac{f(x)}{g(x)} = \text{lim}_{x \rightarrow c} \frac{f'(x)}{g'(x)} \] where \ f'(x) \ and \ g'(x) \ are the derivatives of \ f(x) \ and \ g(x) \, respectively. Just remember: this rule only works if initially you have an indeterminate form.

An indeterminate form arises when the limit of a function is not immediately clear and could take on multiple values. Common indeterminate forms include: \

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• \[ \frac{0}{0} \] \
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• \[ \frac{\text{∞}}{\text{∞}} \] \
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• \[ 0 \times \text{∞} \] \
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• \[ \text{∞} - \text{∞} \] \
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• \[ 0^0 \] \
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• \[ \text{∞}^0 \] \
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• \[ 1^\text{∞} \] \
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To solve an indeterminate form, we often use L'Hôpital's Rule, algebraic manipulation, or other limit properties. In our exercise: \[ \text{lim}_{x \rightarrow 4} \frac{\text{ln}(x / 4)}{x^2 - 16} = \frac{\text{ln}(4/4)}{16 - 16} = \frac{0}{0}, \] which clearly is an indeterminate form of \( \frac{0}{0} \). This hints that L'Hôpital's Rule might be a suitable method to solve it.

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. In the context of applying L'Hôpital's Rule, you need to differentiate the numerator and the denominator independently. For derivatives: \

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• The derivative of \( \text{ln}(x/4) \) is \( \frac{1}{x/4} \times \frac{1}{4} = \frac{1}{x} \). \
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• The derivative of \( x^2 - 16 \) is \( 2x \). \
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Applying L'Hôpital's Rule to our limit: \[ \text{lim}_{x \rightarrow 4} \frac{\text{ln}(x / 4)}{x^2 - 16} = \text{lim}_{x \rightarrow 4} \frac{\frac{1}{x}}{2x} = \text{lim}_{x \rightarrow 4} \frac{1}{2x^2} \]. Now, substitute \( x = 4 \): \[ \frac{1}{2(4)^2} = \frac{1}{32} \]. This gives us the final evaluated limit.