Evaluate :
`lim_(xto 0) (10^(x) - 2^(x) -5^(x)+1)/(sin^(2)x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{\sin^2 x}, \]

we will follow a step-by-step approach.

### Step 1: Rewrite the expression
We start with the limit expression:

\[ \lim_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{\sin^2 x}. \]

### Step 2: Use properties of exponents
We can express \(10^x\) as \(2^x \cdot 5^x\):

\[ 10^x - 2^x - 5^x + 1 = 2^x \cdot 5^x - 2^x - 5^x + 1. \]

### Step 3: Factor the numerator
Now, we can factor out \(2^x\) from the first two terms:

\[ = 2^x (5^x - 1) - (5^x - 1). \]

This can be rewritten as:

\[ = (2^x - 1)(5^x - 1). \]

### Step 4: Substitute back into the limit
Now we substitute this back into our limit:

\[ \lim_{x \to 0} \frac{(2^x - 1)(5^x - 1)}{\sin^2 x}. \]

### Step 5: Apply limit properties
We know from limit properties that:

\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a. \]

Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{(2^x - 1)}{x} \cdot \frac{(5^x - 1)}{x} \cdot \frac{x^2}{\sin^2 x}. \]

### Step 6: Evaluate each limit
Now we can evaluate each part separately:

1. \(\lim_{x \to 0} \frac{2^x - 1}{x} = \ln 2\),
2. \(\lim_{x \to 0} \frac{5^x - 1}{x} = \ln 5\),
3. \(\lim_{x \to 0} \frac{x^2}{\sin^2 x} = 1\).

### Step 7: Combine the results
Combining these results, we have:

\[ \lim_{x \to 0} \frac{(2^x - 1)(5^x - 1)}{\sin^2 x} = \ln 2 \cdot \ln 5 \cdot 1 = \ln 2 \cdot \ln 5. \]

### Final Answer
Thus, the final answer is:

\[ \lim_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{\sin^2 x} = \ln 2 \cdot \ln 5. \]

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