The value of `lim_(xrarr0^(-))(2^(1//x)+2^(3//x))/(3(2^(1//x))+5(2^(3//x))` is

Options

A. `1//3`
B. `1//5`
C. 1
D. `1//4`

Answer

To solve the limit \( \lim_{x \to 0^-} \frac{2^{1/x} + 2^{3/x}}{3 \cdot 2^{1/x} + 5 \cdot 2^{3/x}} \), we will follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the expression in the limit:
\[ \lim_{x \to 0^-} \frac{2^{1/x} + 2^{3/x}}{3 \cdot 2^{1/x} + 5 \cdot 2^{3/x}} \]

### Step 2: Factor out \( 2^{1/x} \)
Next, we factor out \( 2^{1/x} \) from both the numerator and the denominator:
\[ = \lim_{x \to 0^-} \frac{2^{1/x} \left(1 + 2^{2/x}\right)}{2^{1/x} \left(3 + 5 \cdot 2^{2/x}\right)} \]

### Step 3: Cancel \( 2^{1/x} \)
Since \( 2^{1/x} \) is common in both the numerator and the denominator, we can cancel it out (as long as \( 2^{1/x} \neq 0 \)):
\[ = \lim_{x \to 0^-} \frac{1 + 2^{2/x}}{3 + 5 \cdot 2^{2/x}} \]

### Step 4: Analyze \( 2^{2/x} \) as \( x \to 0^- \)
As \( x \to 0^- \), \( \frac{2}{x} \to -\infty \), which means \( 2^{2/x} \to 0 \):
\[ = \frac{1 + 0}{3 + 0} = \frac{1}{3} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{1}{3} \]