Question

y = f(x)
y = g(x)
(a) \\(\lim_{x\to2}f(x)+g(x)\\)
(b) \\(\lim_{x\to0}f(x)-g(x)\\)
(c) \\(\lim_{x\to - 1}f(x)g(x)\\)
(d) \\(\lim_{x\to3}\frac{f(x)}{g(x)}\\)
(e) \\(\lim_{x\to - 2}x^{2}f(x)\\)
(f) \\(f(-1)+\lim_{x\to1}g(x)\\)

Explanation:

Step1: Recall limit - sum rule

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

Step2: Recall limit - difference rule

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

Step3: Recall limit - product rule

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

Step4: Recall limit - quotient rule

$\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$, provided $\lim_{x
ightarrow a}g(x)
eq0$

Step5: Recall limit - constant - multiple rule

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

Step6: Evaluate $\lim_{x

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

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

ightarrow1}g(x)$
From the graph of $y = f(x)$, $f(-1)=2$. From the graph of $y = g(x)$, $\lim_{x
ightarrow1}g(x)=1$. So $f(-1)+\lim_{x
ightarrow1}g(x)=2 + 1=3$

Answer:

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