find the limit.\nlim(x→2) √(x² + 2x + 1)\ndoes not exist\n3\n9\n±3

Explanation:

Step1: Simplify the expression inside square - root

The expression $x^{2}+2x + 1$ is a perfect - square trinomial. Using the formula $(a + b)^2=a^{2}+2ab + b^{2}$, where $a=x$ and $b = 1$, we have $x^{2}+2x + 1=(x + 1)^{2}$. So, $\lim_{x\rightarrow2}\sqrt{x^{2}+2x + 1}=\lim_{x\rightarrow2}\sqrt{(x + 1)^{2}}$.

Step2: Evaluate the limit

The square - root function $y=\sqrt{u}$ is continuous for $u\geq0$. When $x\rightarrow2$, we substitute $x = 2$ into $\sqrt{(x + 1)^{2}}$. Since $\sqrt{(x + 1)^{2}}=\vert x + 1\vert$, and when $x = 2$, $\vert2+1\vert=3$. So, $\lim_{x\rightarrow2}\sqrt{x^{2}+2x + 1}=3$.

Answer:

B. 3