Questions: lim as x approaches infinity of (sqrt(4-7x))/(4x^3-9) i. DNE ii. -4 / 9 iii. 0 iv. -7 / 4

Solution Steps

To find the limit as \( x \) approaches infinity for the given expression, we need to analyze the behavior of both the numerator and the denominator. The numerator is a square root function, and the denominator is a polynomial. As \( x \) approaches infinity, the dominant term in the denominator will be \( 4x^3 \). We should simplify the expression by dividing both the numerator and the denominator by the highest power of \( x \) present in the denominator, which is \( x^3 \).

We need to evaluate the limit \( \lim_{x \rightarrow \infty} \frac{\sqrt{4 - 7x}}{4x^3 - 9} \). As \( x \) approaches infinity, the term \( -7x \) in the square root will dominate, leading to a negative value under the square root, which suggests that the expression will not yield a real number.

To analyze the limit more rigorously, we can rewrite the expression by dividing both the numerator and the denominator by \( x^3 \): \[ \lim_{x \rightarrow \infty} \frac{\sqrt{4 - 7x}}{4x^3 - 9} = \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{4 - 7x}}{x^{3/2}}}{4 - \frac{9}{x^3}} \] As \( x \) approaches infinity, \( \frac{9}{x^3} \) approaches 0, and we focus on the behavior of the numerator.

In the numerator, \( \sqrt{4 - 7x} \) approaches \( \sqrt{-7x} \) as \( x \) becomes very large, which is not defined in the real number system. Therefore, the limit does not exist in the real number sense.

Final Answer