To solve the limit [tex]\(\lim _{x \rightarrow \infty} \frac{4 x^2 + 3 x + 5}{7 x^2 + 2 x + 1}\)[/tex], follow these steps:
1. Identify the highest power of [tex]\( x \)[/tex] in both the numerator and the denominator:
In this function, the highest power of [tex]\( x \)[/tex] is [tex]\( x^2 \)[/tex].
2. Divide every term in the numerator and the denominator by [tex]\( x^2 \)[/tex]:
[tex]\[
\frac{4 x^2 + 3 x + 5}{7 x^2 + 2 x + 1} = \frac{\frac{4 x^2}{x^2} + \frac{3 x}{x^2} + \frac{5}{x^2}}{\frac{7 x^2}{x^2} + \frac{2 x}{x^2} + \frac{1}{x^2}}
\][/tex]
3. Simplify each term:
[tex]\[
= \frac{4 + \frac{3}{x} + \frac{5}{x^2}}{7 + \frac{2}{x} + \frac{1}{x^2}}
\][/tex]
4. Evaluate the limit as [tex]\( x \)[/tex] approaches infinity:
As [tex]\( x \)[/tex] approaches infinity, the terms [tex]\( \frac{3}{x} \)[/tex], [tex]\( \frac{5}{x^2} \)[/tex], [tex]\( \frac{2}{x} \)[/tex], and [tex]\( \frac{1}{x^2} \)[/tex] all approach 0. Thus, the fraction simplifies to:
[tex]\[
\frac{4 + 0 + 0}{7 + 0 + 0} = \frac{4}{7}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim _{x \rightarrow \infty} \frac{4 x^2 + 3 x + 5}{7 x^2 + 2 x + 1} = \frac{4}{7} \approx 0.5714285714285714
\][/tex]
