Question

math 2413: worksheet 2 c. $limlimits_{x \to infty} \frac{2x + 3}{x^2 + 4x + 3}$

Explanation:

Step1: Divide numerator and denominator by \(x^2\)

To find the limit as \(x\to\infty\), we divide each term in the numerator and the denominator by the highest power of \(x\) in the denominator, which is \(x^2\).
So we have:
\[
\lim_{x\to\infty}\frac{\frac{2x}{x^2}+\frac{3}{x^2}}{\frac{x^2}{x^2}+\frac{4x}{x^2}+\frac{3}{x^2}}
\]
Simplifying each term:
\[
\lim_{x\to\infty}\frac{\frac{2}{x}+\frac{3}{x^2}}{1 + \frac{4}{x}+\frac{3}{x^2}}
\]

Step2: Evaluate the limit as \(x\to\infty\)

As \(x\to\infty\), the terms \(\frac{2}{x}\), \(\frac{3}{x^2}\), and \(\frac{4}{x}\) all approach \(0\) because the denominator of each fraction becomes very large, making the whole fraction approach \(0\).
So substituting the limits of each term:
\[
\frac{0 + 0}{1+0 + 0}
\]

Step3: Simplify the expression

Simplifying the numerator and the denominator:
\[
\frac{0}{1}=0
\]

Answer:

\(0\)