The value of the limit $\lim_{x \to 0} (\frac{x^x + 2^x + 3^x + 4^x}{4})^{1/x}$ is
- A. $1$
- B. $3!^{1/3!}$
- C. $3!^{1/4}$
- D. $4!^{1/4}$
✅ Correct Answer: $4!^{1/4}$
Explanation
Using the standard result $\lim_{x \to 0} (\frac{a^x + b^x + c^x + d^x}{4})^{1/x} = (abcd)^{1/4}$ Substituting $a=1, b=2, c=3, d=4$ $(1 \cdot 2 \cdot 3 \cdot 4)^{1/4} = (4!)^{1/4}$
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