Let `lim _(xtooo) (2 ^(x) +a ^(x) +e ^(x))^(1//x)=L` which of the following statement (s) is (are) correct ?

Options

A. if `L =a (a gt 0),` then the range of a is `[e,oo)`
B. if `L=2e(a gt 0),` then the range of a is `{2e}`
C. if `L =e (a gt 0),` then the range of a is `(0,e]`
D. if `L =2a (a gt 1),` then the range of a is `((e)/(2), oo)`

Answer

To solve the limit problem given in the question, we need to evaluate the limit:

\[ L = \lim_{x \to \infty} (2^x + a^x + e^x)^{\frac{1}{x}} \]

### Step 1: Identify the dominant term
As \( x \) approaches infinity, the term with the highest base will dominate the sum. We need to compare \( 2^x \), \( a^x \), and \( e^x \).

- If \( a < 2 \), then \( 2^x \) will dominate.
- If \( a = 2 \), then \( 2^x \) and \( a^x \) are equal.
- If \( a > 2 \), then \( a^x \) will dominate.
- If \( a < e \), then \( e^x \) will dominate.
- If \( a = e \), then \( e^x \) and \( a^x \) are equal.
- If \( a > e \), then \( a^x \) will dominate.

### Step 2: Rewrite the limit based on the dominant term
Assuming \( a \) is the largest of the three bases, we can factor \( a^x \) out of the expression:

\[ L = \lim_{x \to \infty} a^{\frac{1}{x}} \left( \left( \frac{2^x}{a^x} + 1 + \frac{e^x}{a^x} \right) \right)^{\frac{1}{x}} \]

### Step 3: Simplify the limit
Now, we simplify the expression inside the limit:

\[ L = \lim_{x \to \infty} a^{\frac{1}{x}} \left( \left( \left( \frac{2}{a} \right)^x + 1 + \left( \frac{e}{a} \right)^x \right) \right)^{\frac{1}{x}} \]

As \( x \to \infty \):
- If \( a > 2 \), then \( \frac{2}{a} < 1 \) and \( \frac{e}{a} < 1 \), leading to:

\[ L = \lim_{x \to \infty} a^{\frac{1}{x}} \cdot 1 = 1 \]

- If \( a = 2 \), then:

\[ L = \lim_{x \to \infty} 2^{\frac{1}{x}} \cdot 2 = 2 \]

- If \( a < 2 \), then \( \frac{2}{a} > 1 \) and \( L \) will diverge.

### Step 4: Conclusion
Thus, we can conclude:
- If \( a > 2 \), \( L = a \).
- If \( a = 2 \), \( L = 2 \).
- If \( a < 2 \), \( L \) diverges.