hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)\nscore: 4/6 answered: 5/6\n× question 5\nscore…

hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)\nscore: 4/6 answered: 5/6\n× question 5\nscore on last try: 0 of 1 pts. see details for more.\n> next question get a similar question you can retry this question below\nevaluate the limit\n\\(\\lim_{x\\to+\\infty}\\frac{\\sqrt{7 + 2x^{2}}}{9+11x}\\)

hw6 limits at infinity and asymptotes (target l3; §2.2,4.6)\nscore: 4/6 answered: 5/6\n× question 5\nscore on last try: 0 of 1 pts. see details for more.\n> next question get a similar question you can retry this question below\nevaluate the limit\n\\(\\lim_{x\\to+\\infty}\\frac{\\sqrt{7 + 2x^{2}}}{9+11x}\\)

Answer

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\rightarrow+\infty}\frac{\sqrt{7 + 2x^{2}}}{9 + 11x}=\lim_{x\rightarrow+\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\rightarrow+\infty}\frac{\sqrt{\frac{7}{x^{2}}+2}}{\frac{9}{x}+11}).

Step3: Evaluate the limit of each term

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