The value of `lim_(xrarr0^(-))(4^((3)/(x))+15(2^((1)/(x))))/(2^(1+(6)/(x))+6(2^((1)/(x))))` is equal to

Answer

To solve the limit problem, we need to evaluate the expression:

\[ \lim_{x \to 0^-} \frac{4^{\frac{3}{x}} + 15 \cdot 2^{\frac{1}{x}}}{2^{1 + \frac{6}{x}} + 6 \cdot 2^{\frac{1}{x}}} \]

### Step 1: Substitute \( x \) with \( -h \)

Since we are taking the limit as \( x \) approaches \( 0 \) from the left, we substitute \( x = -h \) where \( h \to 0^+ \).

\[ \lim_{h \to 0^+} \frac{4^{\frac{3}{-h}} + 15 \cdot 2^{\frac{1}{-h}}}{2^{1 + \frac{6}{-h}} + 6 \cdot 2^{\frac{1}{-h}}} \]

### Step 2: Simplify the Exponents

Using the property \( a^{-b} = \frac{1}{a^b} \):

\[ = \lim_{h \to 0^+} \frac{\frac{1}{4^{\frac{3}{h}}} + 15 \cdot \frac{1}{2^{\frac{1}{h}}}}{2^{1 - \frac{6}{h}} + 6 \cdot \frac{1}{2^{\frac{1}{h}}}} \]

### Step 3: Rewrite the Terms

Now, rewrite the terms in the limit:

\[ = \lim_{h \to 0^+} \frac{\frac{1}{(2^2)^{\frac{3}{h}}} + 15 \cdot \frac{1}{2^{\frac{1}{h}}}}{2^{1 - \frac{6}{h}} + 6 \cdot \frac{1}{2^{\frac{1}{h}}}} \]

This simplifies to:

\[ = \lim_{h \to 0^+} \frac{\frac{1}{2^{\frac{6}{h}}} + 15 \cdot \frac{1}{2^{\frac{1}{h}}}}{2^{1 - \frac{6}{h}} + 6 \cdot \frac{1}{2^{\frac{1}{h}}}} \]

### Step 4: Factor Out Dominant Terms

As \( h \to 0^+ \), \( 2^{\frac{1}{h}} \) and \( 2^{\frac{6}{h}} \) grow very large. Thus, we factor out the dominant terms:

\[ = \lim_{h \to 0^+} \frac{\frac{1}{2^{\frac{6}{h}}}(1 + 15 \cdot 2^{\frac{5}{h}})}{2^{1 - \frac{6}{h}}(1 + 6 \cdot 2^{\frac{5}{h}})} \]

### Step 5: Cancel Out Terms

The limit can now be simplified:

\[ = \lim_{h \to 0^+} \frac{1 + 15 \cdot 2^{\frac{5}{h}}}{2^{1} \cdot (1 + 6 \cdot 2^{\frac{5}{h}})} \]

### Step 6: Evaluate the Limit

As \( h \to 0^+ \), \( 2^{\frac{5}{h}} \to \infty \):

\[ = \frac{15 \cdot \infty}{2 \cdot 6 \cdot \infty} = \frac{15}{12} = \frac{5}{4} \]

### Final Answer

Thus, the value of the limit is:

\[ \lim_{x \to 0^-} \frac{4^{\frac{3}{x}} + 15 \cdot 2^{\frac{1}{x}}}{2^{1 + \frac{6}{x}} + 6 \cdot 2^{\frac{1}{x}}} = 2.5 \]