Question

The value of \( \lim_{x \to 0} \left(\frac{e^x + 2^x + 4^x}{3}\right)^{\frac{2}{x}} \) is:

Hint:

Use \(a^x \approx 1 + x\ln a\) and \((1+ax)^{b/x} = e^{ab}\).

Answer 1

Correct Answer: 16

Step 1: Understanding the Concept:
The limit is in the indeterminate form \(1^{\infty}\). We use the standard transformation for such limits.
Step 2: Key Formula or Approach:
If \(\lim g(x) = 1\) and \(\lim h(x) = \infty\), then \(\lim [g(x)]^{h(x)} = e^{\lim (g(x) - 1) h(x)}\).
Step 3: Detailed Explanation:
1. Set up the exponential form:
\[ \text{Limit} = e^{\lim_{x\to 0} \left( \frac{8^{x} + 2^{x} + 4^{x}}{3} - 1 \right) \cdot \frac{2}{x}} \]
\[ = e^{2 \cdot \lim_{x\to 0} \frac{8^{x} + 2^{x} + 4^{x} - 3}{3x}} \]
2. Rearrange the numerator:
\[ = e^{\frac{2}{3} \cdot \lim_{x\to 0} \left( \frac{8^{x}-1}{x} + \frac{2^{x}-1}{x} + \frac{4^{x}-1}{x} \right)} \]
3. Use the standard limit \(\lim_{x \to 0} \frac{a^{x}-1}{x} = \ln a\):
\[ = e^{\frac{2}{3} [\ln 8 + \ln 2 + \ln 4]} \]
4. Use properties of logarithms:
\[ = e^{\frac{2}{3} \ln (8 \times 2 \times 4)} = e^{\frac{2}{3} \ln (64)} \]
\[ = e^{\ln (64^{2/3})} \]
5. Simplify the power:
\[ 64^{2/3} = (4^{3})^{2/3} = 4^{2} = 16 \]
\[ \text{Limit} = 16 \]
Step 4: Final Answer:
The limit is 16.