`lim_(x to infty) sqrt(x^(2)-1)/(2x+1)`=

Options

A. 1
B. 0
C. `-1`
D. `1//2`

Answer

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

### Step 1: Simplify the expression inside the limit
We start with the limit:
\[ \lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1} \]

### Step 2: Factor out \(x\) from the square root
Notice that as \(x\) approaches infinity, \(x^2\) dominates \(-1\) in the square root. We can factor \(x^2\) out of the square root:
\[ \sqrt{x^2 - 1} = \sqrt{x^2(1 - \frac{1}{x^2})} = x\sqrt{1 - \frac{1}{x^2}} \]
Thus, the limit becomes:
\[ \lim_{x \to \infty} \frac{x\sqrt{1 - \frac{1}{x^2}}}{2x + 1} \]

### Step 3: Simplify the denominator
Now, we can simplify the denominator \(2x + 1\):
\[ 2x + 1 = 2x(1 + \frac{1}{2x}) \]
So we rewrite the limit:
\[ \lim_{x \to \infty} \frac{x\sqrt{1 - \frac{1}{x^2}}}{2x(1 + \frac{1}{2x})} \]

### Step 4: Cancel \(x\) in the numerator and denominator
We can cancel \(x\) from the numerator and denominator:
\[ \lim_{x \to \infty} \frac{\sqrt{1 - \frac{1}{x^2}}}{2(1 + \frac{1}{2x})} \]

### Step 5: Evaluate the limit as \(x\) approaches infinity
As \(x\) approaches infinity, \(\frac{1}{x^2}\) approaches \(0\) and \(\frac{1}{2x}\) also approaches \(0\). Therefore:
\[ \sqrt{1 - \frac{1}{x^2}} \to \sqrt{1} = 1 \]
and
\[ 2(1 + \frac{1}{2x}) \to 2(1 + 0) = 2 \]
Thus, the limit simplifies to:
\[ \frac{1}{2} \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1} = \frac{1}{2} \]

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