Calculus Solutions Ex#1.2

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Q.01

Solution:

Given that

$\lim_{x \rightarrow a} f(x)=2,$

$\lim_{x \rightarrow a} g(x)=-4$

$\lim_{x \rightarrow a} h(x)=0.$

(a)

$\begin{aligned} \lim_{x \rightarrow a}[f(x)+2 g(x)] & =\lim_{x \rightarrow a} f(x)+2 \underset{x \rightarrow a}{\operatorname{lim}} g(x) \\ & =2+2(-4)=2-8=-6 \end{aligned}$

(b)

$\begin{aligned} \lim _{x \rightarrow a}[h(x)-3 g(x)+1] & =\lim_{x \rightarrow a} h(x)-3 \underset{x \rightarrow a}{\operatorname{lim}} g(x)+\underset{x \rightarrow a}{\operatorname{lim}}(1) \\ & =0-3(-4)+1=13 \end{aligned}$

(c)

$\lim_{x \rightarrow a}[f(x) g(x)]=\left(\lim _{x \rightarrow a} f(x)\right)\left(\lim_{x \rightarrow a} g(x)\right)$

$=(2)(-4)=-8$

(d)

$\lim_{x \rightarrow a}[g(x)]^2 \Rightarrow\left[\lim_{x \rightarrow a} g(x)\right]^2=[-4]^2=16$

(e)

$\begin{aligned} \lim_{x \rightarrow a} \sqrt[3]{6+f(x)} & =\sqrt[3]{{\lim}_{x \rightarrow a}(6)+\lim_{x \rightarrow a} f(x)} \\ & =\sqrt[3]{6+2}=\sqrt[3]{8}=\left(2^3\right)^{1 / 3}=2 \end{aligned}$

(f)

$\lim _{x \rightarrow a} \frac{2}{g(x)}=\frac{2}{\lim_{x \rightarrow a} g(x)}=\frac{2}{-4}=-1 / 2$

Q.02

Solution:

First of all, we shall find the following limits.

$\lim_{x \rightarrow 0} f(x), \quad \lim_{x \rightarrow 2} f(x), \quad \lim_{x \rightarrow 0} g(x), \quad \lim_{x \rightarrow 2} g(x).$

(i) Limit of $f(x)$ at $x=0$ .

$\lim_{x \rightarrow 0^{-}} f(x)=1$
This is because as $x$ approaches 0 from left side, $f(x)$ approaches 1.

$\lim_{x \rightarrow 0^{+}} f(x)=-2$
The reason is as $x$ approaches 0 from the right side, $f(x)$ approaches -2.

Since $\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x).$

Therefore $\lim_{x \rightarrow 0} f(x)$ does not exist.

(ii) Limit of $f(x)$ at $x=2$

$\lim_{x \rightarrow 2^{-}} f(x)=0$

We see from the figure given in the book that as $x \rightarrow 2$ from left side, $f(x)$ approaches 0.

$\lim_{x \rightarrow 2^{+}} f(x)=0$

because as $x \rightarrow 2$ from the right side, $f(x)$ approaches 0.

Since $\lim_{x \rightarrow 2^{-}} f(x) =0= \lim_{x \rightarrow 2^{+}} f(x).$

Therefore $\lim_{x \rightarrow 2} f(x)=0.$

(iii) Limit of $g(x)$ at $x=2$

$\lim_{x \rightarrow 2^{-}} g(x)=0$

We see from the figure given in the book that as $x \rightarrow 2$ from left side, $g(x)$ approaches 0.

$\lim_{x \rightarrow 2^{+}} g(x)=0$

because as $x \rightarrow 2$ from the right side, $f(x)$ approaches 0.

Since $\lim_{x \rightarrow 2^{-}} g(x) =0= \lim_{x \rightarrow 2^{+}} g(x).$

Therefore $\lim_{x \rightarrow 2} g(x)=0.$

(iv) Limit of $g(x)$ at $x=0$

$\lim_{x \rightarrow 0^{-}} g(x)=2, \quad \quad \lim_{x \rightarrow 0^{+}} g(x)=2.$

As $\quad \lim_{x \rightarrow 0^{-}} g(x)=2=\lim_{x \rightarrow 0^{+}} g(x)$

$\therefore \quad \lim _{x \rightarrow 0} g(x)=2$

(a)

$\lim_{x \rightarrow 2}[f(x)+g(x)]=\lim_{x \rightarrow 2} f(x)+\underset{x \rightarrow 2}{\operatorname{lim}} g(x)=0+0=0$

(b)

$\lim_{x \rightarrow 0}[f(x)+g(x)]=\lim_{x \rightarrow 0} f(x)+\lim_{x \rightarrow 0} g(x)$

limit does not exist because $\lim_{x \rightarrow 0} f(x)$ does not exist.

(c)

$\begin{aligned} \lim_{x \rightarrow 0^{+}}[f(x)+g(x)] & =\lim_{x \rightarrow 0^{+}} f(x)+\lim_{x \rightarrow 0^{+}} g(x) \\ & =-2+2=0 \end{aligned}$

(d)

$\begin{aligned} \lim_{x \rightarrow 0^{-}}[f(x)+g(x)] & =\lim_{x \rightarrow 0^{-}} f(x)+\lim_{x \rightarrow 0^{-}} g(x) \\ \\ & =1+2=3 \end{aligned}$