find lim x→∞ √(x² + 4x + 1)/(5x + 1).

Explanation:

Step1: Divide numerator and denominator by $x$

For $x>0$, $\sqrt{x^{2}+4x + 1}=x\sqrt{1+\frac{4}{x}+\frac{1}{x^{2}}}$. So, $\lim_{x\rightarrow\infty}\frac{\sqrt{x^{2}+4x + 1}}{5x + 1}=\lim_{x\rightarrow\infty}\frac{x\sqrt{1+\frac{4}{x}+\frac{1}{x^{2}}}}{x(5+\frac{1}{x})}$.

Step2: Simplify the expression

Cancel out the common factor $x$ in the fraction: $\lim_{x\rightarrow\infty}\frac{\sqrt{1+\frac{4}{x}+\frac{1}{x^{2}}}}{5+\frac{1}{x}}$.

Step3: Use limit properties

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{4}{x}=0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$ and $\lim_{x\rightarrow\infty}\frac{1}{x}=0$. Then $\lim_{x\rightarrow\infty}\frac{\sqrt{1+\frac{4}{x}+\frac{1}{x^{2}}}}{5+\frac{1}{x}}=\frac{\sqrt{1 + 0+0}}{5+0}$.

Step4: Calculate the final result

$\frac{\sqrt{1}}{5}=\frac{1}{5}$.

Answer:

$\frac{1}{5}$