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(C) For a function to be continuous at $x=4$,the Left Hand Limit ($L$.$H$.$L$),Right Hand Limit ($R$.$H$.$L$),and the value of the function $f(4)$ mus

Vedclass · Accepted Answer

$a=1, b=-1$

Vedclass · Answer

(C) For a function to be continuous at $x=4$,the Left Hand Limit ($L$.$H$.$L$),Right Hand Limit ($R$.$H$.$L$),and the value of the function $f(4)$ must be equal.$L.H.L = \lim_{x 	o 4^-} f(x) = \lim_{h 	o 0} \frac{(4-h)-4}{|(4-h)-4|} + a = \lim_{h 	o 0} \frac{-h}{|-h|} + a = \lim_{h 	o 0} \frac{-h}{h} + a = -1 + a$$R.H.L = \lim_{x 	o 4^+} f(x) = \lim_{h 	o 0} \frac{(4+h)-4}{|(4+h)-4|} + b = \lim_{h 	o 0} \frac{h}{|h|} + b = \lim_{h 	o 0} \frac{h}{h} + b = 1 + b$$f(4) = a + b$Equating $L.H.L = R.H.L = f(4)$:$-1 + a = 1 + b = a + b$From $-1 + a = a + b$,we get $b = -1$.From $1 + b = a + b$,we get $a = 1$.Thus,the values are $a=1$ and $b=-1$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

The function defined by $f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x-4}{|x-4|} + b, & x > 4 \end{cases}$ is continuous at $x = 4$,if the values of $a$ and $b$ are:

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