find the following limit. if the limit is infinite, enter infinity or -infinity. if the limit is not…

Explanation:

Step1: Divide numerator and denominator by x

When (x\to\infty), we know that for (x>0), (\sqrt{x^{2}} = x). Divide both the numerator and denominator of (\frac{3x + 2}{\sqrt{7x^{2}+1}}) by (x). The numerator becomes (\frac{3x+2}{x}=3+\frac{2}{x}), and the denominator becomes (\frac{\sqrt{7x^{2}+1}}{x}=\sqrt{\frac{7x^{2}+1}{x^{2}}}=\sqrt{7+\frac{1}{x^{2}}}). So the function is (\frac{3+\frac{2}{x}}{\sqrt{7 + \frac{1}{x^{2}}}}).

Step2: Apply the limit rules

We know that (\lim_{x\to\infty}\frac{1}{x}=0) and (\lim_{x\to\infty}\frac{1}{x^{2}} = 0). Using the sum - rule and quotient - rule of limits (\lim_{x\to\infty}\frac{3+\frac{2}{x}}{\sqrt{7+\frac{1}{x^{2}}}}=\frac{\lim_{x\to\infty}(3+\frac{2}{x})}{\lim_{x\to\infty}\sqrt{7+\frac{1}{x^{2}}}}). Since (\lim_{x\to\infty}(3+\frac{2}{x})=\lim_{x\to\infty}3+\lim_{x\to\infty}\frac{2}{x}=3 + 0=3) and (\lim_{x\to\infty}\sqrt{7+\frac{1}{x^{2}}}=\sqrt{\lim_{x\to\infty}(7+\frac{1}{x^{2}})}=\sqrt{7 + 0}=\sqrt{7}).

Answer:

(\frac{3}{\sqrt{7}})