Answer:
Hence, functions in options (b) and (d) are continuous at x= -4.
Step-by-step explanation:
A function f(x) is continuous at x=a; if the left hand limit(L.H.L) at a=right hand limit (R.H.L.) at a=f(a).
(a)
We are given function f(x) as:
[tex]f(x)=\dfrac{x^2-16}{x+4}=\dfrac{(x-4)(x+4)}{x+4}=x-4[/tex] when x≠ -4
and f(x)=0 when x= -4
L.H.L at x= -4 is f(x)= -8 ( since limit x→ -4 f(x)= -8 (as -4-4= -8))
R.H.L. at x=-4 is f(x)= -8
Also f(-4)=0
hence function f(x) is not continuous at x= -4. ( since L.H.L=R.H.L.≠f(-4) )
(b)
We are given function f(x) as:
[tex]f(x)=\dfrac{x^2-16}{x+4}=\dfrac{(x-4)(x+4)}{x+4}=x-4[/tex] when x≠ -4
and f(x)= -8 when x≠ -4.
Clearly as in part (1)
L.H.L at x= -4 is f(x)= -8 ( since limit x→ -4 f(x)= -8 (as -4-4= -8))
R.H.L. at x=-4 is f(x)= -8
Also f(-4)= -8.
Hence, the function f(x) is continuous ( since L.H.L=R.H.L.=f(-4) )
(c)
We are given function f(x) as:
[tex]f(x)=\dfrac{x^2+8x+16}{x+4}=\dfrac{x^2+4x+4x+16}{x+4}=\dfrac{x(x+4)+4(x+4)}{x+4}=\dfrac{(x+4)(x+4)}{x+4}=x+4\\\\f(x)=x+4[/tex] when x≠-4
and f(x)=16 when x= -4
L.H.L. at x= -4 is 0 ( since limit x→ -4 f(x)=0 (as -4+4=0) )
R.H.L. at x= -4 is 0.
f(-4)= 16
Hence, the function f(x) is not continuous. ( since L.H.L=R.H.L.≠f(-4) )
(d)
We are given function f(x) as:
[tex]f(x)=\dfrac{x^2+8x+16}{x+4}=\dfrac{x^2+4x+4x+16}{x+4}=\dfrac{x(x+4)+4(x+4)}{x+4}=\dfrac{(x+4)(x+4)}{x+4}=x+4\\\\f(x)=x+4[/tex] when x≠-4
and f(x)=0 when x= -4
L.H.L. at x= -4 is 0 ( since limit x→ -4 f(x)=0 (as -4+4=0) )
R.H.L. at x= -4 is 0.
f(-4)= 0
Hence, the function f(x) is continuous. ( since L.H.L=R.H.L.=f(-4) )
