Questions: Evaluate the limit. lim as x approaches infinity of (4x^3+1)/(2x^3+sqrt(16x^6+1)) [A] 0 [B] 2 [C] 1/4 [D] infinity [E] 2/3

Solution Steps

To evaluate the limit as \( x \) approaches infinity for the given function, we need to analyze the dominant terms in the numerator and the denominator. The highest power of \( x \) in both the numerator and the denominator will determine the behavior of the function as \( x \) approaches infinity.

To evaluate the limit as \( x \to \infty \), we first identify the dominant terms in the numerator and the denominator. The dominant term in the numerator is \( 4x^3 \) and in the denominator, it is \( \sqrt{16x^6} \).

We simplify the expression by dividing both the numerator and the denominator by \( x^3 \): \[ \lim_{x \to \infty} \frac{4x^3 + 1}{2x^3 + \sqrt{16x^6 + 1}} = \lim_{x \to \infty} \frac{4 + \frac{1}{x^3}}{2 + \sqrt{16 + \frac{1}{x^6}}} \]

As \( x \to \infty \), the terms \( \frac{1}{x^3} \) and \( \frac{1}{x^6} \) approach 0. Therefore, the expression simplifies to: \[ \lim_{x \to \infty} \frac{4 + 0}{2 + \sqrt{16 + 0}} = \frac{4}{2 + 4} = \frac{4}{6} = \frac{2}{3} \]

Final Answer