Calculus Solutions Ex#1.5

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Q.05

Solution:

Consider the function

$f(x)=\left\{\begin{array}{ll} 1, & x \neq 4 \\ -1 & x=4 \end{array} \text { and } g(x)= \begin{cases}4 x-10, & x \neq 4 \\ -6 & x=4\end{cases}\right.$

For each part, is the given function continuous at $x=4$ ?
(a) $f(x)$

$\begin{aligned} &f(x)=\left\{\begin{array}{ll} 1, & x \neq 4 \\ -1 & x=4 \end{array} \\ & \lim _{x \rightarrow 4} f(x)=1 \text { and } f(4)=-1 \\ & \Rightarrow \lim _{x \rightarrow 4} f(x) \neq f(4) \end{aligned}$

$\Rightarrow f$ is not continaous at $x=4$
(b) $g(x)$

$g(x)= \begin{cases}4 x-10, & x \neq 4 \\ -6 & x=4\end{cases}\right.$

$\lim _{x \rightarrow 4} g(x)=\operatorname{lim}_{x \rightarrow 4}(4 x-10)=4(4)-10=6$

and $g(4)=-6$

Since $\lim _{x \rightarrow 4} g(x) \neq g(4)$ $\therefore$ $g$ is not continuous at $x=4$

Q.05(c-d)

Solution:

$g(x)= \begin{cases}4 x-10, & x \neq 4 \\ -6 & x=4\end{cases}\right.$

$\begin{aligned} \lim_{x \rightarrow 4}[-g(x)] & =-\left[\lim _{x \rightarrow 4} g(x)\right] \\ &=-\left[\lim _{x \rightarrow 4} (4x-10)\right] \\ & =-[4(4)-10] \\ & =-[16-10]=-6 \\ \lim_{x \rightarrow 4}[-g(x)] & =-6 \end{aligned}$

and

$\begin{aligned} & -g(4)=-[g(4)]=-[-6]=6 \\ & \lim _{x \rightarrow 4}[-g(x)] \neq-g(4) \end{aligned}$

Therefore, $-g(x)$ in not continuous at $x=4$ .

(d) $|f(x)|$

$\begin{aligned} & f(x)=\left\{\begin{array}{ll} 1, & x \neq 4 \\ -1 & x=4 \end{array} \\ & \lim_{x \rightarrow 4}|f(x)|=\left|\lim _{x \rightarrow 4} f(x)\right|=|1|=1 \\ & \Rightarrow \quad \lim _{x \rightarrow 4}|f(x)|=1 \\ & \text { and }|f(4)|=|-1|=1 \\ \end{aligned}$

So $\lim _{x \rightarrow 4}|f(x)|=1=|f(4)|$ $\therefore|f(x)|$ in continuous at $x=4$ .

Q.05(e-f)

Solution:

(e) $f(x) g(x)$

$f(x)=\left\{\begin{array}{ll} 1, & x \neq 4 \\ -1 & x=4 \end{array}$

$g(x)= \begin{cases}4 x-10, & x \neq 4 \\ -6 & x=4\end{cases}\right.$

$\begin{aligned} & \lim _{x \rightarrow 4} f(x) g(x)=\left[\lim_{x \rightarrow 4} f(x)\right]\left[\lim_{x \rightarrow 4} g(x)\right] \\ &=[1]\left[\lim _{x \rightarrow 4}(4 x-10)\right] \\ &=[1][4(4)-10] \\ & \lim_{x \rightarrow 4} f(x) g(x)=16-10=6 \\ & \text { and } \quad f(4) g(4)=(-1)(-6)=6 \\ & \Rightarrow \lim_{x \rightarrow 4} f(x) g(x)=6=f(4) g(4) \end{aligned}$

$\Rightarrow f(x) g(x)$ is continuous at $x=4$

(f) $\quad g(f(x))$

$f(x)=\left\{\begin{array}{ll} 1, & x \neq 4 \\ -1 & x=4 \end{array}$

$g(x)= \begin{cases}4 x-10, & x \neq 4 \\ -6 & x=4\end{cases}\right.$

As $\quad \quad$ $f(x)=1$ when $x \neq 4$

$\therefore g(f(x))=g(1) \text { when } x \neq 4$

and $f(x)=-1$ at $x=4$

$\begin{aligned} & \quad \therefore g(f(x))=g(-1) \text { when } x=4 \\ & g(x))= \begin{cases}g(1), & x \neq 4 \\ g(-1), & x=4\end{cases} \end{aligned}$

As $\quad \quad$ $g(x)=4 x-10$ , when $x \neq 4$

$\therefore g(1)=4(1)-10=-6$

and $g(-1)=4(-1)-10=-14$ , when $x=4$

$g(f(x))= \begin{cases}-6, & x \neq 4 \\ -14, & x=4\end{cases}$

$\lim _{x \rightarrow 4} g(f(x))=-6 \text { and } g(f(4))=-14$

Since

$\lim _{x \rightarrow 4} g(f(x)) \neq g(f(4))$

$\therefore \quad g(f(x))$ is not continuous at $x=4$

Q.05(g)

Solution:

(g) $g(x)-6 f(x)$

$f(x)=\left\{\begin{array}{ll} 1, & x \neq 4 \\ -1 & x=4 \end{array}$

$g(x)= \begin{cases}4 x-10, & x \neq 4 \\ -6 & x=4\end{cases}\right.$

$\begin{aligned} \lim_{x \rightarrow 4}[g(x)-6 f(x)] & =\lim _{x \rightarrow 4} g(x)-6 \underset{x \rightarrow 4} f(x) \\ & =\lim_{x \rightarrow 4}[4 x-10]-6 \lim_{x \rightarrow 4}(1) \\ & =4(4)-10-6 \\ & =16-10-6=0 \end{aligned}$

$\begin{aligned} & \lim _{x \rightarrow 4}[g(x)-6 f(x)]=0 \\ & g(4)-6 f(4)=-6-6(-1) \\ & =-6+6=0 \end{aligned}$

Since

$\lim _{x \rightarrow 4}[g(x)-6 f(x)]=0=g(4)-6 f(4)$

therefore $g(x)-6 f(x)$ is continuous at $x=4$ .