Evaluate the following limits :
`Lim_(x to 0) (8^(x)-2^(x))/x`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{8^x - 2^x}{x}, \]

we can follow these steps:

### Step 1: Rewrite the expression

We can rewrite the limit by adding and subtracting 1:

\[ \lim_{x \to 0} \frac{8^x - 1 - (2^x - 1)}{x}. \]

This allows us to separate the terms:

\[ \lim_{x \to 0} \left( \frac{8^x - 1}{x} - \frac{2^x - 1}{x} \right). \]

### Step 2: Evaluate the individual limits

Now we can evaluate each limit separately:

1. For the first limit:

\[ \lim_{x \to 0} \frac{8^x - 1}{x}, \]

we know that this limit can be expressed as \(\ln(8)\) using the property of derivatives (specifically, the derivative of \(a^x\) at \(x=0\) is \(\ln(a)\)).

2. For the second limit:

\[ \lim_{x \to 0} \frac{2^x - 1}{x}, \]

similarly, this limit can be expressed as \(\ln(2)\).

### Step 3: Combine the results

Now we can combine the results from both limits:

\[ \lim_{x \to 0} \frac{8^x - 1}{x} - \lim_{x \to 0} \frac{2^x - 1}{x} = \ln(8) - \ln(2). \]

Using the properties of logarithms, we can simplify this:

\[ \ln(8) - \ln(2) = \ln\left(\frac{8}{2}\right) = \ln(4). \]

### Final Result

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{8^x - 2^x}{x} = \ln(4). \]

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