Given,
\(f(x) = \begin{cases} x^2+as+b & \quad \text, 0\leq{x} <{2}\\ 3x+2, & \quad \text, 2\leq{x}\leq{4}\\2ax+5b & \quad \text,4<x\leq{8} \end{cases} \)
To find \(\lim\limits_{x \to 2}f(x)\)
L.H.L
\(=\lim\limits_{x \to 2^-}f(x)\)
\(=\lim\limits_{x \to 2^-}(x^2+ax+b)\)
= 22 + a∙2 + b
= 2a + b + 4
And,
R.H.L
\(=\lim\limits_{x \to 2^+}f(x)\)
\(=\lim\limits_{x \to 2^+}(3x+2)\)
= 3 ∙ 2 + 2 = 8
Since \(\lim\limits_{x \to 2}f(x) \) exists,
∴ \(\lim\limits_{x \to 2^-}f(x)\) \(=\lim\limits_{x \to 2^+}f(x)\)
⇒ 2a + b + 4 = 8
⇒ 2a + b = 4 … (1)
To find \(\lim\limits_{x \to 2}f(x)\).
L.H.L
\(=\lim\limits_{x \to 4^-}f(x)\)
\(=\lim\limits_{x \to 4^-}(3x+2)\)
= 3 ∙ 4 + 2 = 14
And
R.H.L
\(=\lim\limits_{x \to 4^+}f(x)\)
\(=\lim\limits_{x \to 4^+}(2ax+5b)\)
= 2a. 4 + 5b
= 8a + 5b
Since \(\lim\limits_{x \to 4}f(x) \) exists.
∴ \(\lim\limits_{x \to 4^-}f(x) \) = \(\lim\limits_{x \to 4^+}f(x) \)
⇒ 8a + 5b = 14 … (2)
From (1) and (2),
a = 3 and b = -2.
