Question

Evaluate \(\lim_{x \to +\infty} \frac{x}{\sqrt{x^2 + 1000}}\)

• A. 0
• B. 1
• C. \(\infty\)
• D. 1/2

Hint:

For limits at infinity of algebraic functions, you can ignore the lower-power constant terms.
\[ \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1000}} \approx \lim_{x \to \infty} \frac{x}{\sqrt{x^2}} = \lim_{x \to \infty} \frac{x}{x} = 1 \]
This gives the answer in seconds.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires evaluating the limit of an algebraic fraction as the independent variable \(x\) approaches positive infinity.
As \(x \to \infty\), the constant terms under the radical become negligible compared to the \(x^2\) term.

Step 2: Key Formula or Approach:
To evaluate limits at infinity, divide the numerator and denominator by the highest power of \(x\) present in the expression.
Since \(x \to +\infty\), \(x\) is strictly positive, so we can write:
\[ x = \sqrt{x^2} \]

Step 3: Detailed Explanation:
Let us write down the given limit:
\[ L = \lim_{x \to +\infty} \frac{x}{\sqrt{x^2 + 1000}} \]
Divide both the numerator and the denominator by \(x\):
\[ L = \lim_{x \to +\infty} \frac{\frac{x}{x}}{\frac{\sqrt{x^2 + 1000}}{x}} \]
Since \(x > 0\), we rewrite the denominator using \(x = \sqrt{x^2}\):
\[ L = \lim_{x \to +\infty} \frac{1}{\sqrt{\frac{x^2 + 1000}{x^2}}} \]
Simplify the expression inside the square root:
\[ L = \lim_{x \to +\infty} \frac{1}{\sqrt{1 + \frac{1000}{x^2}}} \]
Now, evaluate the limit as \(x\) approaches positive infinity.
As \(x \to \infty\), the term \(\frac{1000}{x^2}\) approaches 0:
\[ \lim_{x \to +\infty} \frac{1000}{x^2} = 0 \]
Substitute this value into our limit expression:
\[ L = \frac{1}{\sqrt{1 + 0}} = \frac{1}{1} = 1 \]
The limit value is 1, which matches Option (B).

Step 4: Final Answer:
Therefore, the correct option is (B).