Question

If \( f(x) = 5x \csc(\sqrt{x}) - (x-2) \csc(\sqrt{x}) \), then \( \lim\limits_{x \to \infty} f(x^2) \) is:

• A. \( 1 \)
• B. \( -1 \)
• C. \( 5 \)
• D. \( -5 \)

Hint:

For rational functions with oscillatory terms, isolate the dominant terms and simplify the limit by dividing by the highest power of \( x \).

Answer 1

The Correct Option is C

We are given the function: \[ f(x) = \frac{5x \csc(\sqrt{x}) - 1}{(x-2) \csc(\sqrt{x})}. \] We need to evaluate: \[ \lim\limits_{x \to \infty} f(x^2). \] Step 1: Substituting \( x^2 \) into \( f(x) \)
\[ f(x^2) = \frac{5x^2 \csc(\sqrt{x^2}) - 1}{(x^2 - 2) \csc(\sqrt{x^2})}. \] Since \( \sqrt{x^2} = x \), we simplify: \[ f(x^2) = \frac{5x^2 \csc(x) - 1}{(x^2 - 2) \csc(x)}. \] Step 2: Evaluating the limit as \( x \to \infty \)
For large \( x \), the behavior of \( \csc(x) \) oscillates but remains finite. Hence, the dominant terms in the numerator and denominator are: \[ 5x^2 \csc(x) \quad {and} \quad x^2 \csc(x). \] Dividing both numerator and denominator by \( x^2 \csc(x) \): \[ \lim\limits_{x \to \infty} f(x^2) = \lim\limits_{x \to \infty} \frac{5x^2 \csc(x) - 1}{(x^2 - 2) \csc(x)} = \lim\limits_{x \to \infty} \frac{5 - \frac{1}{x^2 \csc(x)}}{1 - \frac{2}{x^2}}. \] As \( x \to \infty \), both \( \frac{1}{x^2 \csc(x)} \) and \( \frac{2}{x^2} \) approach zero, leaving: \[ \frac{5 - 0}{1 - 0} = 5. \]