find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the limit does not otherwise…

Explanation:

Step1: Factor out (x^{2}) from the square - root in the numerator

[ \begin{align*} \lim_{x\rightarrow+\infty}\frac{\sqrt{x + 2x^{2}}}{3x-1}&=\lim_{x\rightarrow+\infty}\frac{\sqrt{x^{2}(\frac{1}{x}+2)}}{3x - 1}\ \end{align*} ] Since (x>0) as (x\rightarrow+\infty), (\sqrt{x^{2}}=x). So the expression becomes (\lim_{x\rightarrow+\infty}\frac{x\sqrt{\frac{1}{x}+2}}{3x - 1})

Step2: Divide both numerator and denominator by (x)

[ \begin{align*} \lim_{x\rightarrow+\infty}\frac{x\sqrt{\frac{1}{x}+2}}{3x - 1}&=\lim_{x\rightarrow+\infty}\frac{\sqrt{\frac{1}{x}+2}}{3-\frac{1}{x}} \end{align*} ]

Step3: Use the limit laws (\lim_{x\rightarrow a}(f(x)\pm g(x))=\lim_{x\rightarrow a}f(x)\pm\lim_{x\rightarrow a}g(x)) and (\lim_{x\rightarrow a}(f(x)g(x))=\lim_{x\rightarrow a}f(x)\cdot\lim_{x\rightarrow a}g(x)) ((a = +\infty) here)

We know that (\lim_{x\rightarrow+\infty}\frac{1}{x}=0). Then (\lim_{x\rightarrow+\infty}\sqrt{\frac{1}{x}+2}=\sqrt{\lim_{x\rightarrow+\infty}\frac{1}{x}+2}=\sqrt{0 + 2}=\sqrt{2}) and (\lim_{x\rightarrow+\infty}(3-\frac{1}{x})=\lim_{x\rightarrow+\infty}3-\lim_{x\rightarrow+\infty}\frac{1}{x}=3-0 = 3)

Answer:

(\frac{\sqrt{2}}{3})