Answer
a) $\lim\limits_{x \to 2}[f(x)+5g(x)]=-6$
b) $\lim\limits_{x \to 2}[g(x)]^3=-8$
c) $\lim\limits_{x \to 2}\sqrt{f(x)}=2$
d) $\lim\limits_{x \to 2}\frac{3f(x)}{g(x)}=-6$
e) $\lim\limits_{x \to 2}\frac{g(x)}{h(x)}$ does not exist
f) $\lim\limits_{x \to 2}\frac{g(x)h(x)}{f(x)}=0$
Work Step by Step
a) $\lim\limits_{x \to 2}[f(x) + 5g(x)]=\lim\limits_{x \to 2}f(x)+5\lim\limits_{x \to 2}g(x)=4+5\times(-2)=-6$
b) $\lim\limits_{x \to 2}[g(x)]^3=[\lim\limits_{x \to 2}g(x)]^3=(-2)^3=-8$
c) $\lim\limits_{x \to 2}\sqrt {f(x)}=\sqrt{\lim\limits_{x \to 2}f(x)}=\sqrt4=2$
d) $\lim\limits_{x \to 2}\frac{3f(x)}{g(x)}=\frac{3\times\lim\limits_{x \to 2}f(x)}{\lim\limits_{x \to 2}g(x)}=\frac{3\times4}{-2}=-6$
e) $\lim\limits_{x \to 2}\frac{g(x)}{h(x)}=\frac{\lim\limits_{x \to 2}g(x)}{\lim\limits_{x \to 2}h(x)}$
However, we notice that $\lim\limits_{x \to 2}h(x)=0$
Therefore, this limit is undefined and as a result, does not exist.
f) $\lim\limits_{x \to 2}\frac{g(x)h(x)}{f(x)}=\frac{\lim\limits_{x \to 2}g(x)\times\lim\limits_{x \to 2}h(x)}{\lim\limits_{x \to 2}f(x)}=\frac{(-2)\times 0}{4}=0$
