Question

hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)
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evaluate the limit
\\(\lim_{x\to+\infty}\frac{\sqrt{7 + 2x^{2}}}{9+11x}\\)

Explanation:

Step1: Divide numerator and denominator by x

For \(x>0\), \(\sqrt{7 + 2x^{2}}=\sqrt{x^{2}(\frac{7}{x^{2}}+2)}=x\sqrt{\frac{7}{x^{2}}+2}\). Then \(\lim_{x
ightarrow+\infty}\frac{\sqrt{7 + 2x^{2}}}{9 + 11x}=\lim_{x
ightarrow+\infty}\frac{x\sqrt{\frac{7}{x^{2}}+2}}{x(\frac{9}{x}+11)}\).

Step2: Simplify the expression

Cancel out the common factor \(x\) in the fraction, we get \(\lim_{x
ightarrow+\infty}\frac{\sqrt{\frac{7}{x^{2}}+2}}{\frac{9}{x}+11}\).

Step3: Evaluate the limit of each term

As \(x
ightarrow+\infty\), \(\lim_{x
ightarrow+\infty}\frac{7}{x^{2}} = 0\) and \(\lim_{x
ightarrow+\infty}\frac{9}{x}=0\). So \(\lim_{x
ightarrow+\infty}\frac{\sqrt{\frac{7}{x^{2}}+2}}{\frac{9}{x}+11}=\frac{\sqrt{0 + 2}}{0+11}\).

Step4: Calculate the final result

\(\frac{\sqrt{2}}{11}\).

Answer:

\(\frac{\sqrt{2}}{11}\)