The value of `lim_(xrarr0^(-))(4^(2+(3)/(x))+5(2^((1)/(x))))/(2^((1+(6)/(x)))+6(2^((1)/(x))))` is equal to

Options

A. `(5)/(6)`
B. 8
C. 16
D. `(5)/(2)`

Answer

To solve the limit problem

\[ \lim_{x \to 0^-} \frac{4^{2 + \frac{3}{x}} + 5 \cdot 2^{\frac{1}{x}}}{2^{1 + \frac{6}{x}} + 6 \cdot 2^{\frac{1}{x}}} \]

we will break it down step by step.

### Step 1: Rewrite the terms in the limit

We start by rewriting \(4\) as \(2^2\):

\[ 4^{2 + \frac{3}{x}} = (2^2)^{2 + \frac{3}{x}} = 2^{4 + \frac{6}{x}} \]

Thus, the limit becomes:

\[ \lim_{x \to 0^-} \frac{2^{4 + \frac{6}{x}} + 5 \cdot 2^{\frac{1}{x}}}{2^{1 + \frac{6}{x}} + 6 \cdot 2^{\frac{1}{x}}} \]

### Step 2: Factor out the dominant term

As \(x\) approaches \(0\) from the left, \( \frac{1}{x} \) approaches \(-\infty\). Therefore, \(2^{\frac{1}{x}}\) approaches \(0\) and \(2^{\frac{6}{x}}\) approaches \(+\infty\).

We can factor out \(2^{\frac{6}{x}}\) from both the numerator and the denominator:

\[ \lim_{x \to 0^-} \frac{2^{\frac{6}{x}} \left(2^4 + 5 \cdot 2^{\frac{1}{x} - \frac{6}{x}}\right)}{2^{\frac{6}{x}} \left(2^1 + 6 \cdot 2^{\frac{1}{x} - \frac{6}{x}}\right)} \]

This simplifies to:

\[ \lim_{x \to 0^-} \frac{2^4 + 5 \cdot 2^{\frac{1}{x} - \frac{6}{x}}}{2^1 + 6 \cdot 2^{\frac{1}{x} - \frac{6}{x}}} \]

### Step 3: Analyze the limit as \(x \to 0^-\)

As \(x \to 0^-\), \( \frac{1}{x} - \frac{6}{x} = -\frac{5}{x} \to +\infty\). Thus, \(2^{\frac{1}{x} - \frac{6}{x}} \to +\infty\).

Now the limit becomes:

\[ \lim_{x \to 0^-} \frac{2^4 + 5 \cdot \infty}{2^1 + 6 \cdot \infty} \]

This simplifies to:

\[ \lim_{x \to 0^-} \frac{5 \cdot \infty}{6 \cdot \infty} = \frac{5}{6} \]

### Final Result

Thus, the value of the limit is:

\[ \frac{5}{6} \]