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hw6 limits at infinity and asymptotes (target l3; §2.2,4,6)
score: 0/6 answered: 0/6
question 1
evaluate the limit
\\(\lim_{x\to\infty}\frac{2 + 3x}{6-4x}\\)

Explanation:

Step1: Divide numerator and denominator by x

Divide each term in the fraction $\frac{2 + 3x}{6-4x}$ by $x$. We get $\lim_{x
ightarrow\infty}\frac{\frac{2}{x}+3}{\frac{6}{x}-4}$.

Step2: Evaluate limits of individual terms

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{2}{x}=0$ and $\lim_{x
ightarrow\infty}\frac{6}{x}=0$. So we have $\frac{0 + 3}{0-4}$.

Step3: Simplify the fraction

$\frac{3}{-4}=-\frac{3}{4}$.

Answer:

$-\frac{3}{4}$