Question

hw 4 - limits to infinity section 1.3: problem (1 point) find the following limit. notes: enter dne if the limit does not exist. \\(\lim_{x\to+\infty} \frac{4x^2 + 3x + 10}{-5x^2 + 3x + 3} = \square\\)

Explanation:

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

To find the limit as \(x\to+\infty\) of a rational function, we divide each term in the numerator and the denominator by the highest power of \(x\) in the denominator, which is \(x^2\) here.
For the numerator: \(\frac{4x^2 + 3x + 10}{x^2}=\frac{4x^2}{x^2}+\frac{3x}{x^2}+\frac{10}{x^2}=4+\frac{3}{x}+\frac{10}{x^2}\)
For the denominator: \(\frac{- 5x^2+3x + 3}{x^2}=\frac{-5x^2}{x^2}+\frac{3x}{x^2}+\frac{3}{x^2}=- 5+\frac{3}{x}+\frac{3}{x^2}\)
So the function becomes \(\lim_{x\to+\infty}\frac{4 + \frac{3}{x}+\frac{10}{x^2}}{-5+\frac{3}{x}+\frac{3}{x^2}}\)

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

We know that \(\lim_{x\to+\infty}\frac{1}{x}=0\) and \(\lim_{x\to+\infty}\frac{1}{x^2}=0\) (by the property that for any positive integer \(n\), \(\lim_{x\to\pm\infty}\frac{1}{x^n}=0\)).
Substituting these limits into the numerator and the denominator:
The numerator limit: \(\lim_{x\to+\infty}(4+\frac{3}{x}+\frac{10}{x^2})=4 + 0+0 = 4\)
The denominator limit: \(\lim_{x\to+\infty}(-5+\frac{3}{x}+\frac{3}{x^2})=-5 + 0+0=-5\)

Step3: Find the limit of the fraction

Using the quotient rule for limits (if \(\lim_{x\to a}f(x)=L\) and \(\lim_{x\to a}g(x)=M
eq0\), then \(\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}\)), we have:
\(\lim_{x\to+\infty}\frac{4+\frac{3}{x}+\frac{10}{x^2}}{-5+\frac{3}{x}+\frac{3}{x^2}}=\frac{\lim_{x\to+\infty}(4+\frac{3}{x}+\frac{10}{x^2})}{\lim_{x\to+\infty}(-5+\frac{3}{x}+\frac{3}{x^2})}=\frac{4}{-5}=-\frac{4}{5}\)

Answer:

\(-\frac{4}{5}\)