Continuity

If $f\left(x\right)=\begin{cases} \dfrac{\sin x - \cos x}{x - \frac{\pi}{4}}, & x \neq \dfrac{\pi}{4} \\ k, & x = \dfrac{\pi}{4} \end{cases}$ $\;\;$ is continuous at $x=\dfrac{\pi}{4}$, find k.

Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{4}$ $\implies$ $f\left(\dfrac{\pi}{4}\right) = \lim\limits_{x \to \frac{\pi}{4}} f\left(x\right)$

i.e. $k = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\sin x - \cos x}{x - \frac{\pi}{4}}$

Let $x = \dfrac{\pi}{4}+h$

As $x \to \dfrac{\pi}{4}, \; h \to 0$

$\begin{aligned} \therefore \; k & = \lim\limits_{h \to 0} \; \dfrac{\sin \left(\dfrac{\pi}{4}+h\right)-\cos \left(\dfrac{\pi}{4}+h\right)}{h} \\ & \\ & \left[\begin{aligned} \text{Note: } & \sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ & \cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \end{aligned}\right] \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\sin \left(\dfrac{\pi}{4}\right) \cos h + \cos \left(\dfrac{\pi}{4}\right)\sin h- \left[\cos \left(\dfrac{\pi}{4}\right) \cos h - \sin \left(\dfrac{\pi}{4} \right) \sin h\right]}{h} \\ & \\ & = \lim\limits_{h \to 0} \; \dfrac{\dfrac{\cos h}{\sqrt{2}} + \dfrac{\sin h}{\sqrt{2}} - \dfrac{\cos h}{\sqrt{2}}+ \dfrac{\sin h}{\sqrt{2}}}{h} \\ & \\ & = \dfrac{2}{\sqrt{2}} \; \lim\limits_{h \to 0} \; \dfrac{\sin h}{h} \\ & \\ & = \sqrt{2} \times 1 = \sqrt{2} \end{aligned}$

Continuity

If $f\left(x\right)= \begin{cases} \dfrac{\sin 3x}{7x} + \alpha, & x > 0 \\ x+3-\beta, & x < 0 \end{cases}$ $\;\;$ is continuous at $x=0$, find $\alpha + \beta$

Since $f\left(x\right)$ is continuous at $x=0$, $\lim\limits_{x \to 0^-} f\left(x\right) = \lim\limits_{x \to 0^+}f\left(x\right)$

i.e. $\lim\limits_{x \to 0^-} \; \left(x+3-\beta\right) = \lim\limits_{x \to 0^+} \; \left(\dfrac{\sin 3x}{7x}+\alpha\right)$

i.e. $3-\beta = \dfrac{3}{7} \; \lim\limits_{x \to 0^+} \; \dfrac{\sin 3x}{3x}+ \lim\limits_{x \to 0^+} \alpha$

i.e. $3-\beta = \dfrac{3}{7} \times 1 + \alpha$

i.e. $\alpha + \beta = 3-\dfrac{3}{7}=\dfrac{18}{7}$

Continuity

Let $f\left(x\right)= \begin{cases} x+a\sqrt{2} \sin x, & 0\leq x < \dfrac{\pi}{4} \\ & \\ 2x \cot x +b, & \dfrac{\pi}{4}\leq x < \dfrac{\pi}{2} \\ & \\ a \cos 2x - b \sin x, & \dfrac{\pi}{2}\leq x \leq \pi \end{cases}$

If $f\left(x\right)$ is continuous for $0 \leq x \leq \pi$, find a and b.

Since $f\left(x\right)$ is continuous for $0 \leq x \leq \pi$, therefore $f\left(x\right)$ must be continuous at $\dfrac{\pi}{4}$ and $\dfrac{\pi}{2}$

$\therefore$ $f\left(\dfrac{\pi}{4}\right) = \lim\limits_{x \to \frac{\pi}{4}^-} f\left(x\right) = \lim\limits_{x \to \frac{\pi}{4}^+} f\left(x\right)$ $\;\;$ $\cdots$ (1)

and $f\left(\dfrac{\pi}{2}\right) = \lim\limits_{x \to \frac{\pi}{2}^-} f\left(x\right) = \lim\limits_{x \to \frac{\pi}{2}^+} f\left(x\right)$ $\;\;$ $\cdots$ (2)

$\therefore$ Equation (1) gives

$2 \times \dfrac{\pi}{4} \times \cot \left(\dfrac{\pi}{4}\right)+b = \lim\limits_{x \to \frac{\pi}{4}^-} \; x+a\sqrt{2} \sin x = \lim\limits_{x \to \frac{\pi}{4}^+} \; 2x \cot x + b$

i.e. $\dfrac{\pi}{2}+b= \dfrac{\pi}{4}+a \sqrt{2} \sin \left(\dfrac{\pi}{4}\right)= 2 \times \dfrac{\pi}{4} \times \cot \left(\dfrac{\pi}{4}\right)+b$

i.e. $\dfrac{\pi}{2}+b= \dfrac{\pi}{4}+a\sqrt{2}\times \dfrac{1}{\sqrt{2}}=\dfrac{\pi}{2}+b$

i.e. $\dfrac{\pi}{4}+a = \dfrac{\pi}{2}+b$

i.e. $a-b=\dfrac{\pi}{4}$ $\cdots$ (3)

Equation (2) gives

$a \cos \left(2 \times \dfrac{\pi}{2}\right) - b\sin \left(\dfrac{\pi}{2}\right) = \lim\limits_{x \to \frac{\pi}{2}^-} \; 2x \cot x +b = \lim\limits_{x \to \frac{\pi}{2}^+} \; a \cos 2x - b\sin x$

i.e. $-a-b = 2 \times \dfrac{\pi}{2} \times \cot \left(\dfrac{\pi}{2}\right)+b = a \cos \left(2 \times \dfrac{\pi}{2}\right)- b\sin\left(\dfrac{\pi}{2}\right)$

i.e. $-a-b = b = -a-b$

i.e. $-a-b = b$

i.e. $a = -2b$ $\cdots$ (4)

Substituting the value of a from equation (4) in equation (3) gives

$-2b-b=\dfrac{\pi}{4}$ $\implies$ $b=-\dfrac{\pi}{12}$

$\therefore$ From equation (4), $a=2 \times \dfrac{\pi}{12}=\dfrac{\pi}{6}$

Continuity

Discuss the continuity of the function $f\left(x\right)= \begin{cases} \dfrac{x-\left|x\right|}{x}, & x \neq 0 \\ 2, & x=0 \end{cases}$ $\;$ at $x=0$

When $x>0, \; \left|x\right|=+x$

When $x<0, \; \left|x\right|=-x$

$\lim\limits_{x \to 0^+}f\left(x\right)=\lim\limits_{x \to 0} \; \dfrac{x-x}{x}=0$

$\lim\limits_{x \to 0^-}f\left(x\right)=\lim\limits_{x \to 0} \; \dfrac{x+x}{x}=\dfrac{2x}{x}=2$

$f\left(0\right)=2$

Since $\lim\limits_{x \to 0^-}f\left(x\right)=f\left(0\right) \neq \lim\limits_{x \to 0^+}f\left(x\right)$, function $f\left(x\right)$ is discontinuous at $x=0$.

Continuity

Find $\alpha$ and $\beta$ so that the function $f\left(x\right)$ defined by $f\left(x\right) = \begin{cases} -2 \sin x, & \text{for } -\pi \leq x \leq - \dfrac{\pi}{2} \\\\ \alpha \sin x + \beta, & \text{for } - \dfrac{\pi}{2} < x < \dfrac{\pi}{2} \\\\ \cos x, & \text{for } \dfrac{\pi}{2} \leq x \leq \pi \end{cases}$ $\;$ is continuous on $\left[-\pi,\pi\right]$

Since $f\left(x\right)$ is continuous on $\left[-\pi, \pi\right]$, therefore $f\left(x\right)$ must be continuous at $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$.

$\therefore$ $f\left(-\dfrac{\pi}{2}\right)= \lim\limits_{x \to -\frac{\pi}{2}^-} f\left(x\right) = \lim\limits_{x \to -\frac{\pi}{2}^+} f\left(x\right)$ $\cdots$ (1)

and $f\left(\dfrac{\pi}{2}\right) = \lim\limits_{x \to \frac{\pi}{2}^-}f\left(x\right)= \lim\limits_{x \to \frac{\pi}{2}^+}f\left(x\right)$ $\cdots$ (2)

$\therefore$ From equation (1) we have

$-2 \sin \left(-\dfrac{\pi}{2}\right) = \lim\limits_{x \to -\frac{\pi}{2}^-} \; -2\sin x = \lim\limits_{x \to -\frac{\pi}{2}^+} \; \alpha \sin x + \beta$

i.e. $-2 \sin \left(-\dfrac{\pi}{2}\right) = -2 \sin \left(-\dfrac{\pi}{2}\right) = \alpha \sin \left(-\dfrac{\pi}{2}\right)+ \beta$

i.e. $2 \sin \left(\dfrac{\pi}{2}\right) = 2 \sin \left(\dfrac{\pi}{2}\right) = -\alpha \sin \left(\dfrac{\pi}{2}\right)+\beta$ $\;\;\; \left[\text{Note: }\sin \left(-\theta\right)=-\sin \theta\right]$

i.e. $-\alpha + \beta = 2 \; \cdots (3)$

From equation (2) we have

$\cos \left(\dfrac{\pi}{2}\right) = \lim\limits_{x \to \frac{\pi}{2}^-} \; \alpha \sin x + \beta = \lim\limits_{x \to \frac{\pi}{2}^+} \; \cos x$

i.e. $\cos \left(\dfrac{\pi}{2}\right) = \alpha \sin \left(\dfrac{\pi}{2}\right)+\beta = \cos \left(\dfrac{\pi}{2}\right)$

i.e. $\alpha + \beta = 0 $

$\implies$ $\alpha = -\beta \; \cdots (4)$

Substituting the value of $\alpha$ in equation (3) gives

$2 \beta = 2$ $\implies$ $\beta = 1$

Substituting the value of $\beta$ in equation (4) gives

$\alpha = -1$

Continuity

If $f\left(x\right)= \dfrac{1-\tan x}{1-\sqrt{2} \sin x}$ for $x \neq \dfrac{\pi}{4}$, is continuous at $x=\dfrac{\pi}{4}$, find $f\left(\dfrac{\pi}{4}\right)$

$\begin{aligned} \lim\limits_{x \to \frac{\pi}{4}} f\left(x\right) & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{1-\tan x}{1-\sqrt{2}\sin x} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{1-\dfrac{\sin x}{\cos x}}{1-\sqrt{2}\sin x} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\cos x - \sin x}{\cos x \left(1-\sqrt{2}\sin x\right)} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\left(\cos x - \sin x\right)\left(cos x + \sin x\right)\left(1+\sqrt{2}\sin x\right)}{\cos x \left(1-\sqrt{2}\sin x\right) \left(1+\sqrt{2}\sin x\right) \left(\cos x + \sin x\right)} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{\left(cos^2 x - \sin ^2 x\right)\left(1+\sqrt{2}\sin x\right)}{\cos x \left(1-2\sin^2 x\right)\left(\cos x + \sin x\right)} \\ & \\ & = \lim\limits_{x \to \frac{\pi}{4}} \; \dfrac{1+\sqrt{2}\sin x}{\cos x \left(\cos x + \sin x\right)}\\ & \\ & \left[\text{Note: }\cos 2\theta = \cos ^2 \theta - \sin^2 \theta = 1-2\sin^2 \theta\right] \\ & \\ & = \dfrac{1+\sqrt{2}\sin \left(\dfrac{\pi}{4}\right)}{\cos \left(\dfrac{\pi}{4}\right)\left[\cos \left(\dfrac{\pi}{4}\right)+ \sin \left(\dfrac{\pi}{4}\right)\right]} \\ & \\ & = \dfrac{1+\sqrt{2}\times \dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}\times \left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)} = 2 \end{aligned}$

Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{4}$,

$\implies$ $f\left(\dfrac{\pi}{4}\right) = \lim\limits_{x \to \frac{\pi}{4}} f\left(x\right) = 2$

Continuity

Let $f\left(x\right)= \begin{cases} \dfrac{1-\sin^3 x}{3 \cos^2 x} & \text{if } x < \dfrac{\pi}{2} \\ & \\ a & \text{if } x = \dfrac{\pi}{2} \\ & \\ \dfrac{b\left(1-\sin x\right)}{\left(\pi - 2x\right)^2} & \text{if } x > \dfrac{\pi}{2} \end{cases}$
If $f\left(x\right)$ be a continuous function at $x=\dfrac{\pi}{2}$, find a and b.

$\begin{aligned} \lim\limits_{x \to \frac{\pi}{2}^-} f\left(x\right) & = \lim\limits_{x \to \frac{\pi}{2}^-} \; \dfrac{1-\sin^3 x}{3\cos^2 x} \\ & = \lim\limits_{x \to \frac{\pi}{2}^-} \; \dfrac{\left(1-\sin x\right)\left(1+\sin x + \sin^2 x\right)}{3 \left(1-\sin^2 x\right)} \\ & = \lim\limits_{x \to \frac{\pi}{2}^-} \; \dfrac{\left(1-\sin x\right)\left(1+\sin x + \sin^2 x\right)}{3 \left(1+\sin x\right)\left(1-\sin x\right)} \\ & = \lim\limits_{x \to \frac{\pi}{2}^-} \; \dfrac{1+\sin x + \sin^2 x}{3 \left(1+\sin x\right)} \\ & = \dfrac{1+1+1^2}{3\left(1+1\right)} = \dfrac{1}{2} \end{aligned}$

$\lim\limits_{x \to \frac{\pi}{2}^+} f\left(x\right) = \lim\limits_{x \to \frac{\pi}{2}^+} \; \dfrac{b\left(1-\sin x\right)}{\left(\pi - 2x\right)^2}$

Let $x = \dfrac{\pi}{2}+h$

Then, as $x \to \dfrac{\pi}{2}, \; h \to 0$

$\begin{aligned} \therefore \; \lim\limits_{x \to \frac{\pi}{2}^+}f\left(x\right) & = b \;\lim\limits_{h \to 0} \; \dfrac{1-\sin \left(\dfrac{\pi}{2}+h\right)}{\left[\pi - 2\left(\dfrac{\pi}{2}+h\right)\right]^2} \\ & = b\; \lim\limits_{h \to 0} \; \dfrac{1-\cos h}{4h^2} \;\; \left[\text{Note: }\sin \left(\dfrac{\pi}{2}+\theta\right)=\cos \theta\right] \\ & = b \;\lim\limits_{h \to 0} \; \dfrac{2 \sin^2 \left(\dfrac{h}{2}\right)}{4h^2} \;\; \left[\text{Note: }1-\cos 2 \theta = 2 \sin^2 \theta\right] \\ & = \dfrac{b}{2} \; \lim\limits_{h \to 0} \; \left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right]^2 \times \dfrac{1}{4} \\ & = \dfrac{b}{2} \times 1^2 \times \dfrac{1}{4} = \dfrac{b}{8} \end{aligned}$

Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{2}$

$\implies$ $\lim\limits_{x \to \frac{\pi}{2}^-}f\left(x\right)=f\left(\dfrac{\pi}{2}\right)=\lim\limits_{x \to \frac{\pi}{2}^+}f\left(x\right)$

i.e. $\dfrac{1}{2} = a = \dfrac{b}{8}$

$\implies$ $a=\dfrac{1}{2}$ and $b=4$

Continuity

If the function $f\left(x\right) = \dfrac{\log_e \left(1+x\right)-\log_e \left(1-x\right)}{x}$ is continuous at $x=0$, find $f\left(0\right)$.

$\left[\begin{aligned} \text{Note: } & \log_e \left(1+x\right) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots \infty \\ & \log_e \left(1-x\right) = -x - \dfrac{x^2}{2} -\dfrac{x^3}{3} - \dfrac{x^4}{4} - \cdots \infty \end{aligned}\right]$

$\lim\limits_{x \to 0} f\left(x\right)$
$=\lim\limits_{x \to 0} \; \dfrac{\log_e \left(1+x\right)-\log_e \left(1-x\right)}{x}$
$\lim\limits_{x \to 0} \; \dfrac{\left[\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+ \cdots \infty\right)-\left(-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}- \cdots \infty\right)\right]}{x}$
$= \lim\limits_{x \to 0} \; \dfrac{2x + \dfrac{2x^3}{3}+\dfrac{2x^5}{5}+ \cdots \infty}{x}$
$= \lim\limits_{x \to 0} \; \dfrac{2x \left(1+\dfrac{x^2}{3}+\dfrac{x^4}{5}+ \cdots \infty\right)}{x}$
$= 2 \lim\limits_{x \to 0} \; \left(1+\dfrac{x^2}{3}+\dfrac{x^4}{5}+ \cdots \infty\right)$
$= 2 \times 1 = 2$

Since the function $f\left(x\right)$ is continuous at $x=0$, $\lim\limits_{x \to 0} f\left(x\right)=f\left(0\right)$

$\therefore$ $f\left(0\right)=2$

Continuity

Locate the points of discontinuity, if any, for the function $f\left(x\right)= \begin{cases} \dfrac{x^3-64}{x^2-16}, & x \neq 4 \\ 1, & x=4 \end{cases}$

$\begin{aligned} f\left(x\right) & = \begin{cases} \dfrac{x^3-64}{x^2-16}, & x \neq 4 \\ 1, & x=4 \end{cases} \\ & \\ & = \begin{cases} \dfrac{\left(x-4\right)\left(x^2+4x+16\right)}{\left(x+4\right)\left(x-4\right)}, & x \neq 4 \\ 1, & x=4 \end{cases} \\ & \\ & = \begin{cases} \dfrac{x^2+4x+16}{x+4}, & x \neq 4 \\ 1, & x=4 \end{cases} \\ & \\ & = \begin{cases} \dfrac{x^2+4x+16}{x+4}, & x>4 \\ \dfrac{x^2+4x+16}{x+4}, & x<4 \\ 1, & x=4 \end{cases} \end{aligned}$

When $x>4$, $f\left(x\right)=\dfrac{x^2+4x+16}{x+4}$ is continuous (polynomial function).

When $x<4$, $f\left(x\right)=\dfrac{x^2+4x+16}{x+4}$ is continuous (polynomial function).

When $x=4$,

$\text{Left Hand Limit (LHL)} = \lim\limits_{x \to 4^-}f\left(x\right)=\lim\limits_{x \to 4^-} \; \dfrac{x^2+4x+16}{x+4}$

Let $x=4-h$

As $x \to 4^-, \; h \to 0$

$\begin{aligned} \therefore \text{LHL} & =\lim\limits_{h \to 0} \; \dfrac{\left(4-h\right)^2+4\left(4-h\right)+16}{4-h+4} \\ & = \lim\limits_{h \to 0} \; \dfrac{16-8h+h^2+16-4h+16}{8-h} \\ & = \lim\limits_{h \to 0} \; \dfrac{h^2-12h+48}{8-h} \\ & = \dfrac{0-0+48}{8-0} = 6 \end{aligned}$

$\text{Right Hand Limit (RHL)} = \lim\limits_{x \to 4^+}f\left(x\right)=\lim\limits_{x \to 4^+} \; \dfrac{x^2+4x+16}{x+4}$

Let $x=4+h$

As $x \to 4^+, \; h \to 0$

$\begin{aligned} \therefore \text{RHL} & =\lim\limits_{h \to 0} \; \dfrac{\left(4+h\right)^2+4\left(4+h\right)+16}{4+h+4} \\ & = \lim\limits_{h \to 0} \; \dfrac{16+8h+h^2+16+4h+16}{8+h} \\ & = \lim\limits_{h \to 0} \; \dfrac{h^2+12h+48}{8+h} \\ & = \dfrac{0+0+48}{8+0} = 6 \end{aligned}$

$f\left(4\right)=1$

$\therefore$ $\text{LHL}=\text{RHL}\neq f\left(4\right)$

$\implies$ $x=4$ is a point of discontinuity.

Continuity

Find the value of constant k so that the function $f\left(x\right)= \begin{cases} \dfrac{k \cos x}{\pi - 2x}, & \text{if } x \neq \dfrac{\pi}{2} \\\\ 3, & \text{if } x = \dfrac{\pi}{2} \end{cases}$
is continuous at $x=\dfrac{\pi}{2}$

$\lim\limits_{x \to \frac{\pi}{2}} f\left(x\right)=\lim\limits_{x \to \frac{\pi}{2}} \; \dfrac{k \cos x}{\pi - 2x}$

Let $x=\dfrac{\pi}{2}+p$

As $x \to \dfrac{\pi}{2}, \; p \to 0$

$\begin{aligned} \therefore \; \lim\limits_{x \to \frac{\pi}{2}} f\left(x\right) & = \lim\limits_{p \to 0} \; \dfrac{k \cos \left(\dfrac{\pi}{2}+p\right)}{\pi -2 \left(\dfrac{\pi}{2}+p\right)} \\ & = \lim\limits_{p \to 0} \; \dfrac{-k \sin p}{-2p} \;\; \left[\text{Note: }\cos \left(\dfrac{\pi}{2}+\theta\right)=-\sin \theta\right] \\ & = \dfrac{k}{2} \; \lim\limits_{p \to 0} \; \dfrac{\sin p}{p} \\ & = \dfrac{k}{2} \times 1 = \dfrac{k}{2} \;\; \cdots (1) \end{aligned}$

$f\left(\dfrac{\pi}{2}\right)=3$ $\;\;$ $\cdots$ (2)

Since $f\left(x\right)$ is continuous at $x=\dfrac{\pi}{2}$, $\;$ $\lim\limits_{x \to \frac{\pi}{2}}f\left(x\right)=f\left(\dfrac{\pi}{2}\right)$

i.e. $\dfrac{k}{2}=3$ [from equations (1) and (2)]

$\implies$ $k=6$

Continuity

The function $f\left(x\right)$ is defined as

$f\left(x\right)= \begin{cases} x^2 + ax + b, & 0 \leq x < 2 \\ & \\ 3 x + 2, & 2 \leq x \leq 4 \\ & \\ 2 a x + 5 b, & 4 < x \leq 8 \end{cases}$

is continuous in $\left[0,8\right]$. Find the values of a and b.

Since $f\left(x\right)$ is continuous in $\left[0,8\right]$, therefore $f\left(x\right)$ must be continuous at 2 and 4.

$\therefore$ $f\left(2\right)= \lim\limits_{x \to 2^-} f\left(x\right) = \lim\limits_{x \to 2^+} f\left(x\right)$ $\cdots$ (1)

and $f\left(4\right) = \lim\limits_{x \to 4^-}f\left(x\right)= \lim\limits_{x \to 4^+}f\left(x\right)$ $\cdots$ (2)

$\therefore$ From equation (1) we have

$\left(3\times2\right)+2 = \lim\limits_{x \to 2^-} \; x^2+ax+b = \lim\limits_{x \to 2^+} \; 3x+2$

i.e. $8=4+2a+b=8$

i.e. $2a+b=4$ $\cdots$ (3)

From equation (2) we have

$\left(3 \times 4\right)+2 = \lim\limits_{x \to 4^-} \; 3x+2 = \lim\limits_{x \to 4^+} \; 2ax+5b$

i.e. $14=14=8a+5b$

i.e. $8a+5b=14$ $\cdots$ (4)

Multiplying equation (3) with 5 and subtracting from equation (4) gives

$2a=6$ $\implies$ $a=3$

$\therefore$ From equation (3), $b=4-\left(2\times 3\right)=-2$

Continuity

For what value(s) of a and b, $f\left(x\right)= \begin{cases} \dfrac{x+2}{\left|x+2\right|}+a, & x<-2 \\ & \\ a+b, & x=-2 \\ & \\ \dfrac{x+2}{\left|x+2\right|}+b, & x>-2 \end{cases}$
is continuous at $x=-2$

$f\left(-2\right)=a+b$ $\cdots$ (1)

When $x<-2, \; \left|x+2\right|=-\left(x+2\right)$

When $x>-2, \; \left|x+2\right|=\left(x+2\right)$

$\begin{aligned} \therefore \lim\limits_{x \to -2^-} f\left(x\right) & =\lim\limits_{x\to -2^-} \; \dfrac{x+2}{\left|x+2\right|}+a \\ & \\ & = \lim\limits_{x \to -2} \; \dfrac{x+2}{-\left(x+2\right)}+a \\ & \\ & = -1+a \;\;\; \cdots (2) \end{aligned}$

$\begin{aligned} \lim\limits_{x \to -2^+} f\left(x\right) & =\lim\limits_{x\to -2^+} \; \dfrac{x+2}{\left|x+2\right|}+b \\ & \\ & = \lim\limits_{x \to -2} \; \dfrac{x+2}{\left(x+2\right)}+b \\ & \\ & = 1+b \;\;\; \cdots (3) \end{aligned}$

Since the function is continuous at $x=-2$,

$\lim\limits_{x \to -2^-}f\left(x\right)=f\left(-2\right)=\lim\limits_{x \to -2^+}f\left(x\right)$

i.e. $-1+a=a+b$ $\implies$ $b=-1$ [from equations (1) and (2)]

and $1+b=a+b$ $\implies$ $a=1$ [from equations (1) and (3)]