What is `lim_(xtooo) (sqrt(a^(2)x^(2)ax+1)sqrt(a^(2)x^(2)+1))` equal to?

Options

A. `(1)/(2)`
B. 1
C. 2
D. 0

Explanation

To solve the limit problem \( \lim_{x \to \infty} \left( \sqrt{a^2 x^2 + ax + 1} - \sqrt{a^2 x^2 + 1} \right) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ L = \lim_{x \to \infty} \left( \sqrt{a^2 x^2 + ax + 1} - \sqrt{a^2 x^2 + 1} \right) \]

### Step 2: Rationalize the expression
To simplify this expression, we can multiply and divide by the conjugate:
\[ L = \lim_{x \to \infty} \frac{\left( \sqrt{a^2 x^2 + ax + 1} - \sqrt{a^2 x^2 + 1} \right) \left( \sqrt{a^2 x^2 + ax + 1} + \sqrt{a^2 x^2 + 1} \right)}{\sqrt{a^2 x^2 + ax + 1} + \sqrt{a^2 x^2 + 1}} \]
This gives:
\[ L = \lim_{x \to \infty} \frac{(a^2 x^2 + ax + 1) - (a^2 x^2 + 1)}{\sqrt{a^2 x^2 + ax + 1} + \sqrt{a^2 x^2 + 1}} \]

### Step 3: Simplify the numerator
The numerator simplifies to:
\[ ax + 1 - 1 = ax \]
Thus, we have:
\[ L = \lim_{x \to \infty} \frac{ax}{\sqrt{a^2 x^2 + ax + 1} + \sqrt{a^2 x^2 + 1}} \]

### Step 4: Factor out \(x\) from the square roots
Next, we can factor \(x^2\) out of the square roots in the denominator:
\[ L = \lim_{x \to \infty} \frac{ax}{\sqrt{x^2(a^2 + \frac{a}{x} + \frac{1}{x^2})} + \sqrt{x^2(a^2 + \frac{1}{x^2})}} \]
This simplifies to:
\[ L = \lim_{x \to \infty} \frac{ax}{x\left(\sqrt{a^2 + \frac{a}{x} + \frac{1}{x^2}} + \sqrt{a^2 + \frac{1}{x^2}}\right)} \]
Cancelling \(x\) gives:
\[ L = \lim_{x \to \infty} \frac{a}{\sqrt{a^2 + \frac{a}{x} + \frac{1}{x^2}} + \sqrt{a^2 + \frac{1}{x^2}}} \]

### Step 5: Apply the limit
As \(x \to \infty\), \(\frac{a}{x} \to 0\) and \(\frac{1}{x^2} \to 0\):
\[ L = \frac{a}{\sqrt{a^2 + 0 + 0} + \sqrt{a^2 + 0}} = \frac{a}{\sqrt{a^2} + \sqrt{a^2}} = \frac{a}{a + a} = \frac{a}{2a} = \frac{1}{2} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to \infty} \left( \sqrt{a^2 x^2 + ax + 1} - \sqrt{a^2 x^2 + 1} \right) = \frac{1}{2} \]

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