drag each tile to the correct box. not all tiles will be used. arrange the expressions for which limits…

Explanation:

Step1: Find limit of $\lim_{x\rightarrow\infty}\frac{3x + 12}{x^{2}+14x + 28}$

As $x\rightarrow\infty$, for a rational - function $\frac{f(x)}{g(x)}$ where $\text{deg}(f)=1$ and $\text{deg}(g)=2$, we use the fact that $\lim_{x\rightarrow\infty}\frac{3x + 12}{x^{2}+14x + 28}=\lim_{x\rightarrow\infty}\frac{\frac{3}{x}+\frac{12}{x^{2}}}{1+\frac{14}{x}+\frac{28}{x^{2}}}=0$.

Step2: Find limit of $\lim_{x\rightarrow3}\frac{x^{2}-5x + 6}{x - 3}$

Factor the numerator: $x^{2}-5x + 6=(x - 2)(x - 3)$. Then $\lim_{x\rightarrow3}\frac{x^{2}-5x + 6}{x - 3}=\lim_{x\rightarrow3}\frac{(x - 2)(x - 3)}{x - 3}=\lim_{x\rightarrow3}(x - 2)=1$.

Step3: Find limit of $\lim_{x\rightarrow - 4}\frac{x^{2}+x - 12}{x + 4}$

Factor the numerator: $x^{2}+x - 12=(x + 4)(x - 3)$. Then $\lim_{x\rightarrow - 4}\frac{x^{2}+x - 12}{x + 4}=\lim_{x\rightarrow - 4}(x - 3)=-7$.

Step4: Find limit of $\lim_{x\rightarrow7}\frac{x^{2}-5x - 14}{3x - 21}$

Factor the numerator: $x^{2}-5x - 14=(x - 7)(x+2)$. Then $\lim_{x\rightarrow7}\frac{x^{2}-5x - 14}{3x - 21}=\lim_{x\rightarrow7}\frac{(x - 7)(x + 2)}{3(x - 7)}=\lim_{x\rightarrow7}\frac{x + 2}{3}=3$.

Step5: Find limit of $\lim_{x\rightarrow5}\frac{x^{2}-5x}{x - 5}$

Factor the numerator: $x^{2}-5x=x(x - 5)$. Then $\lim_{x\rightarrow5}\frac{x^{2}-5x}{x - 5}=\lim_{x\rightarrow5}x = 5$.

Step6: Find limit of $\lim_{x\rightarrow\infty}\frac{2x^{2}+5x - 2}{x^{2}+2}$

Divide both numerator and denominator by $x^{2}$: $\lim_{x\rightarrow\infty}\frac{2x^{2}+5x - 2}{x^{2}+2}=\lim_{x\rightarrow\infty}\frac{2+\frac{5}{x}-\frac{2}{x^{2}}}{1+\frac{2}{x^{2}}}=2$.

Step7: Find limit of $\lim_{x\rightarrow - 2}\frac{x^{2}-11}{x + 2}$

As $x\rightarrow - 2$, the numerator approaches $(-2)^{2}-11=-7$ and the denominator approaches $0$. The limit does not exist.

Step8: Find limit of $\lim_{x\rightarrow - 2}\frac{2x^{2}-5x + 4}{(x + 2)^{2}}$

As $x\rightarrow - 2$, the numerator approaches $2(-2)^{2}-5(-2)+4=8 + 10+4 = 22$ and the denominator approaches $0$ from the positive - side. The limit is $\infty$.

Answer:

$\lim_{x\rightarrow - 4}\frac{x^{2}+x - 12}{x + 4},\lim_{x\rightarrow\infty}\frac{3x + 12}{x^{2}+14x + 28},\lim_{x\rightarrow\infty}\frac{2x^{2}+5x - 2}{x^{2}+2},\lim_{x\rightarrow3}\frac{x^{2}-5x + 6}{x - 3},\lim_{x\rightarrow7}\frac{x^{2}-5x - 14}{3x - 21},\lim_{x\rightarrow5}\frac{x^{2}-5x}{x - 5}$