Question

math 151 calculus i quiz 1
name: key
note: show all your work to earn full credit. ta:

1. (8 points) use the graph to determine the following:

(a) \\(\lim_{x\to - 2}f(x)=0\\)
(b) \\(\lim_{x\to - 1}f(x)= - 2\\)
(c) \\(\lim_{x\to1}f(x)=0\\)
(d) \\(\lim_{x\to3}f(x)=dne\\)
(e) \\(\lim_{x\to4}f(x)=1\\)
(f) \\(f(2)=0\\)

1. (9 points) suppose \\(\lim_{x\to4}f(x)=2\\) and \\(\lim_{x\to4}g(x)= - 3\\). find

(a) \\(\lim_{x\to4}g(x)-5f(x)\\)
\\(=\lim_{x\to4}g(x)-5\lim_{x\to4}f(x)\\)
\\(=-3 - 5\cdot2\\)
\\(=-3 - 10\\)
\\(=-13\\)
(b) \\(\lim_{x\to4}g(x)^{2}\\)
\\(=\lim_{x\to4}g(x)^{2}\\)
\\(=(-3)^{2}\\)
\\(=9\\)
(c) \\(\lim_{x\to4}\frac{g(x)}{f(x)-1}=\frac{\lim_{x\to4}g(x)}{\lim_{x\to4}f(x)-1}=\frac{-3}{2 - 1}=-3\\)

Explanation:

Step1: Recall limit - sum rule

For $\lim_{x
ightarrow a}[g(x)-5f(x)]$, by the sum - difference rule of limits $\lim_{x
ightarrow a}(u(x)\pm v(x))=\lim_{x
ightarrow a}u(x)\pm\lim_{x
ightarrow a}v(x)$. So $\lim_{x
ightarrow a}[g(x)-5f(x)]=\lim_{x
ightarrow a}g(x)-5\lim_{x
ightarrow a}f(x)$. Given $\lim_{x
ightarrow a}f(x) = 2$ and $\lim_{x
ightarrow a}g(x)=-3$, we have $\lim_{x
ightarrow a}g(x)-5\lim_{x
ightarrow a}f(x)=-3 - 5\times2=-3 - 10=-13$.

Step2: Recall limit - power rule

For $\lim_{x
ightarrow a}[g(x)]^{2}$, by the power rule of limits $\lim_{x
ightarrow a}[u(x)]^{n}=[\lim_{x
ightarrow a}u(x)]^{n}$. Since $\lim_{x
ightarrow a}g(x)=-3$, then $\lim_{x
ightarrow a}[g(x)]^{2}=(-3)^{2}=9$.

Step3: Recall limit - quotient rule

For $\lim_{x
ightarrow a}\frac{g(x)}{f(x)-1}$, by the quotient rule of limits $\lim_{x
ightarrow a}\frac{u(x)}{v(x)}=\frac{\lim_{x
ightarrow a}u(x)}{\lim_{x
ightarrow a}v(x)}$ (where $\lim_{x
ightarrow a}v(x)
eq0$). Here $\lim_{x
ightarrow a}g(x)=-3$ and $\lim_{x
ightarrow a}(f(x)-1)=\lim_{x
ightarrow a}f(x)-1$. Since $\lim_{x
ightarrow a}f(x) = 2$, then $\lim_{x
ightarrow a}(f(x)-1)=2 - 1=1$. So $\lim_{x
ightarrow a}\frac{g(x)}{f(x)-1}=\frac{-3}{2 - 1}=-3$.

Answer:

(a) - 13
(b) 9
(c) - 3