_x infinity (4x2 - 5x + 1)/(sqrt(9x4 + 2x2 - 6))

Question

ScanSolve AI Tutor · Accepted Answer

Here is the solution to the limit problem.

We need to evaluate the limit:

\lim_{x \to \infty} \frac{4x^2 - 5x + 1}{\sqrt{9x^4 + 2x^2 - 6}}

Identify the highest power of $x$ in the denominator.
The highest power of $x$ inside the square root is $x^4$ . When $x \to \infty$ , $\sqrt{x^4} = x^2$ . Therefore, we will divide every term in the numerator and the denominator by $x^2$ . For terms inside the square root, we divide by $x^4$ .
Divide each term by the identified highest power of $x$ .
$\lim_{x \to \infty} \frac{\frac{4x^2}{x^2} - \frac{5x}{x^2} + \frac{1}{x^2}}{\sqrt{\frac{9x^4}{x^4} + \frac{2x^2}{x^4} - \frac{6}{x^4}}}$
Simplify the terms.
$\lim_{x \to \infty} \frac{4 - \frac{5}{x} + \frac{1}{x^2}}{\sqrt{9 + \frac{2}{x^2} - \frac{6}{x^4}}}$
Apply the limit as $x \to \infty$ .
As $x \to \infty$ , any term of the form $\frac{c}{x^n}$ (where $c$ is a constant and $n > 0$ ) approaches 0.
$\frac{4 - 0 + 0}{\sqrt{9 + 0 - 0}}$
Simplify the expression.
$\frac{4}{\sqrt{9}} = \frac{4}{3}$
The limit is $\frac{4}{3}$ .

Answer