Question

question 1 (limlimits_{x \to infty} \frac{6x + 4}{11x^2 - 5x + 11}) (\bigcirc) 4/11 (\bigcirc) dne (\bigcirc) 6/11 (\bigcirc) 0

Explanation:

Step1: Analyze degrees of numerator and denominator

The numerator \(6x + 4\) is a first - degree polynomial (the highest power of \(x\) is \(1\)), and the denominator \(11x^{2}-5x + 11\) is a second - degree polynomial (the highest power of \(x\) is \(2\)). When finding the limit as \(x
ightarrow\infty\) of a rational function \(\frac{f(x)}{g(x)}\), if the degree of the numerator \(n\) is less than the degree of the denominator \(m\) (\(nightarrow\infty}\frac{f(x)}{g(x)} = 0\). Here, \(n = 1\) and \(m=2\), so \(n

Step2: Confirm the limit

We can also use the method of dividing both the numerator and the denominator by the highest power of \(x\) in the denominator. The highest power of \(x\) in the denominator is \(x^{2}\).
Divide the numerator \(6x + 4\) by \(x^{2}\): \(\frac{6x + 4}{x^{2}}=\frac{6}{x}+\frac{4}{x^{2}}\)
Divide the denominator \(11x^{2}-5x + 11\) by \(x^{2}\): \(\frac{11x^{2}-5x + 11}{x^{2}}=11-\frac{5}{x}+\frac{11}{x^{2}}\)
Now, find the limit as \(x
ightarrow\infty\):
\(\lim_{x
ightarrow\infty}\frac{6x + 4}{11x^{2}-5x + 11}=\lim_{x
ightarrow\infty}\frac{\frac{6}{x}+\frac{4}{x^{2}}}{11-\frac{5}{x}+\frac{11}{x^{2}}}\)
We know that \(\lim_{x
ightarrow\infty}\frac{1}{x}=0\) and \(\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0\).
Substitute these limits into the expression:
\(\frac{\lim_{x
ightarrow\infty}(\frac{6}{x}+\frac{4}{x^{2}})}{\lim_{x
ightarrow\infty}(11-\frac{5}{x}+\frac{11}{x^{2}})}=\frac{0 + 0}{11-0 + 0}=0\)

Answer:

\(0\)