y = f(x)\ny = g(x)\n(a) \\(\\lim_{x\\to2}f(x)+g(x)\\)\n(b) \\(\\lim_{x\\to0}f(x)-g(x)\\)\n(c)…

Explanation:

Step1: Recall limit - sum rule

$\lim_{x\rightarrow a}[f(x)+g(x)]=\lim_{x\rightarrow a}f(x)+\lim_{x\rightarrow a}g(x)$

Step2: Recall limit - difference rule

$\lim_{x\rightarrow a}[f(x)-g(x)]=\lim_{x\rightarrow a}f(x)-\lim_{x\rightarrow a}g(x)$

Step3: Recall limit - product rule

$\lim_{x\rightarrow a}[f(x)g(x)]=\lim_{x\rightarrow a}f(x)\cdot\lim_{x\rightarrow a}g(x)$

Step4: Recall limit - quotient rule

$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}$, provided $\lim_{x\rightarrow a}g(x)\neq0$

Step5: Recall limit - constant - multiple rule

$\lim_{x\rightarrow a}[cf(x)] = c\lim_{x\rightarrow a}f(x)$

Step6: Evaluate $\lim_{x\rightarrow - 2}[x^{2}f(x)]$

By the product - rule $\lim_{x\rightarrow - 2}[x^{2}f(x)]=\lim_{x\rightarrow - 2}x^{2}\cdot\lim_{x\rightarrow - 2}f(x)$. We know that $\lim_{x\rightarrow - 2}x^{2}=(-2)^{2} = 4$. From the graph of $y = f(x)$, $\lim_{x\rightarrow - 2}f(x)=1$. So $\lim_{x\rightarrow - 2}[x^{2}f(x)]=4\times1 = 4$

Step7: Evaluate $f(-1)+\lim_{x\rightarrow1}g(x)$

From the graph of $y = f(x)$, $f(-1)=2$. From the graph of $y = g(x)$, $\lim_{x\rightarrow1}g(x)=1$. So $f(-1)+\lim_{x\rightarrow1}g(x)=2 + 1=3$

Answer:

(a) 1 (b) DNE (c) 2 (d) DNE (e) 4 (f) 3