Question

divide the numerator and denominator by the highest power of x in the denominator.
lim_{x
ightarrowinfty}\frac{x - 4}{sqrt{4x^{2}+1}}
lim_{x
ightarrowinfty}\frac{x - 4}{sqrt{4x^{2}+1}}=\frac{1}{2}\text{ (simplify your answer.)}

Explanation:

Step1: Identify highest - power of x

The highest power of $x$ in the denominator $\sqrt{4x^{2}+1}$ is $x$ (since $\sqrt{4x^{2}} = 2|x|$ as $x\to\infty$). Divide numerator and denominator by $x$.
$\lim_{x\to\infty}\frac{x - 4}{\sqrt{4x^{2}+1}}=\lim_{x\to\infty}\frac{\frac{x}{x}-\frac{4}{x}}{\frac{\sqrt{4x^{2}+1}}{x}}$

Step2: Simplify denominator

We know that $\frac{\sqrt{4x^{2}+1}}{x}=\sqrt{\frac{4x^{2}+1}{x^{2}}}=\sqrt{4+\frac{1}{x^{2}}}$ for $x>0$. So the limit becomes $\lim_{x\to\infty}\frac{1-\frac{4}{x}}{\sqrt{4 + \frac{1}{x^{2}}}}$.

Step3: Evaluate limit

As $x\to\infty$, $\frac{4}{x}\to0$ and $\frac{1}{x^{2}}\to0$. Then $\lim_{x\to\infty}\frac{1-\frac{4}{x}}{\sqrt{4+\frac{1}{x^{2}}}}=\frac{1 - 0}{\sqrt{4+0}}$.

Step4: Calculate final result

$\frac{1 - 0}{\sqrt{4+0}}=\frac{1}{2}$.

Answer:

$\frac{1}{2}$