Evaluate the limit $\lim _ { x \rightarrow 4 ^ { + } } ( 8 ( x - 4 ) ) ^ { x - 4 }$ using L'Hopital's Rule if necessary.
A) 1
B) $e$
C) 0
D) $e ^ { 4 }$
E) $e ^ { 8 }$

Question

Evaluate the limit $\lim _ { x \rightarrow 4 ^ { + } } ( 8 ( x - 4 ) ) ^ { x - 4 }$ using L'Hopital's Rule if necessary. 
A) 1
B) $e$
C) 0
D) $e ^ { 4 }$
E) $e ^ { 8 }$

Quizplus · Accepted Answer

The expression can be rewritten as $e^{\ln((8(x-4))^{x-4})}$. Applying the limit, we get $e^{\lim _ { x \rightarrow 4 ^ { + } }(x-4)\ln(8(x-4))}$. This is an indeterminate form of $0 \cdot (-\infty)$, so we can use L'Hopital's Rule by rewriting it as $\frac{\ln(8(x-4))}{\frac{1}{x-4}}$. Differentiating the numerator and denominator separately gives us $\frac{8}{8(x-4)} \cdot -1 = \frac{-1}{x-4}$, and applying the limit as $x$ approaches 4 from the right gives $-\infty$. However, the correct approach to solve this limit is to recognize that as $x$ approaches 4, the base $8(x-4)$ approaches 0 and the exponent $x-4$ also approaches 0, leading to the form $0^0$, which is indeterminate. However, this specific limit approaches 1, not through direct application of L'Hopital's Rule, but through understanding the properties of exponential functions and their limits. My explanation mistakenly applied L'Hopital's Rule in an incorrect manner for this specific problem. The correct approach involves recognizing the limit of the form $0^0$ and understanding that in this context, it simplifies to 1.

Evaluate the Limit lim⁡x→4+(8(x−4))x−4\lim _ { x \rightarrow 4 ^ { + } } ( 8 ( x - 4 ) ) ^ { x - 4 }limx→4+​(8(x−4))x−4

Evaluate the Limit $\lim _ { x \rightarrow 4 ^ { + } } ( 8 ( x - 4 ) ) ^ { x - 4 }$