Question

evaluate the limit

lim_{x→+∞} (11x² - 9x + 10)/(7x + 11)

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Explanation:

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

Divide each term in the numerator \(11x^{2}-9x + 10\) and denominator \(7x + 11\) by \(x\).
The numerator becomes \(\frac{11x^{2}}{x}-\frac{9x}{x}+\frac{10}{x}=11x-9+\frac{10}{x}\), and the denominator becomes \(\frac{7x}{x}+\frac{11}{x}=7+\frac{11}{x}\). So the limit is \(\lim_{x
ightarrow+\infty}\frac{11x - 9+\frac{10}{x}}{7+\frac{11}{x}}\).

Step2: Use the limit properties

We know that \(\lim_{x
ightarrow+\infty}\frac{a}{x}=0\) for any constant \(a\).
\(\lim_{x
ightarrow+\infty}(11x - 9+\frac{10}{x})=\lim_{x
ightarrow+\infty}(11x)-\lim_{x
ightarrow+\infty}(9)+\lim_{x
ightarrow+\infty}(\frac{10}{x})\). Since \(\lim_{x
ightarrow+\infty}(9) = 9\) and \(\lim_{x
ightarrow+\infty}(\frac{10}{x})=0\), and \(\lim_{x
ightarrow+\infty}(11x)=+\infty\).
\(\lim_{x
ightarrow+\infty}(7+\frac{11}{x})=\lim_{x
ightarrow+\infty}(7)+\lim_{x
ightarrow+\infty}(\frac{11}{x}) = 7\) (because \(\lim_{x
ightarrow+\infty}(\frac{11}{x})=0\)).
So \(\lim_{x
ightarrow+\infty}\frac{11x - 9+\frac{10}{x}}{7+\frac{11}{x}}=\frac{\lim_{x
ightarrow+\infty}(11x)-9 + 0}{7+0}\). As \(x
ightarrow+\infty\), the value of the fraction \(\frac{11x-9+\frac{10}{x}}{7+\frac{11}{x}}\) is dominated by the term \(11x\) in the numerator.

Answer:

\(+\infty\)