Question

evaluate the limit
\\( \lim _ { x \
ightarrow \infty } \frac { 6 x + 11 } { 2 x ^ { 2 } - 6 x + 7 } \\)
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Explanation:

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

$$\lim_{x ightarrow\infty}\frac{\frac{6x}{x^{2}}+\frac{11}{x^{2}}}{\frac{2x^{2}}{x^{2}}-\frac{6x}{x^{2}}+\frac{7}{x^{2}}}=\lim_{x ightarrow\infty}\frac{\frac{6}{x}+\frac{11}{x^{2}}}{2 - \frac{6}{x}+\frac{7}{x^{2}}}$$

Step2: Use the limit property \(\lim_{x

ightarrow\infty}\frac{1}{x^{n}} = 0\) (\(n>0\))
As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{6}{x}=0\), \(\lim_{x
ightarrow\infty}\frac{11}{x^{2}} = 0\), \(\lim_{x
ightarrow\infty}\frac{6}{x}=0\), \(\lim_{x
ightarrow\infty}\frac{7}{x^{2}}=0\)

So \(\lim_{x
ightarrow\infty}\frac{\frac{6}{x}+\frac{11}{x^{2}}}{2-\frac{6}{x}+\frac{7}{x^{2}}}=\frac{0 + 0}{2-0 + 0}\)

Answer:

\(0\)