Question

find the horizontal asymptotes for the following function.
f(x)=\frac{4x + sqrt{x^{2}-4}}{2x - 3}
y=\text{symbolic expression}
y=\text{symbolic expression}

Explanation:

Step1: Consider limit as x approaches positive infinity

Divide numerator and denominator by \(x\) (since \(x>0\) as \(x\to+\infty\)). We have \(\lim_{x\to+\infty}\frac{4x+\sqrt{x^{2}-4}}{2x - 3}=\lim_{x\to+\infty}\frac{4+\sqrt{1-\frac{4}{x^{2}}}}{2-\frac{3}{x}}\).

Step2: Evaluate the limit

As \(x\to+\infty\), \(\frac{4}{x^{2}}\to0\) and \(\frac{3}{x}\to0\). So \(\lim_{x\to+\infty}\frac{4+\sqrt{1-\frac{4}{x^{2}}}}{2-\frac{3}{x}}=\frac{4 + \sqrt{1-0}}{2-0}=\frac{4 + 1}{2}=\frac{5}{2}\).

Step3: Consider limit as x approaches negative infinity

Divide numerator and denominator by \(|x|=-x\) (since \(x<0\) as \(x\to-\infty\)). The function becomes \(\lim_{x\to-\infty}\frac{4x+\sqrt{x^{2}-4}}{2x - 3}=\lim_{x\to-\infty}\frac{4-\sqrt{1-\frac{4}{x^{2}}}}{2-\frac{3}{x}}\).

Step4: Evaluate the limit

As \(x\to-\infty\), \(\frac{4}{x^{2}}\to0\) and \(\frac{3}{x}\to0\). So \(\lim_{x\to-\infty}\frac{4-\sqrt{1-\frac{4}{x^{2}}}}{2-\frac{3}{x}}=\frac{4-\sqrt{1 - 0}}{2-0}=\frac{4 - 1}{2}=\frac{3}{2}\).

Answer:

\(y=\frac{5}{2}\), \(y=\frac{3}{2}\)