We have, f (x) = $\begin{cases} 2x-3, & x \le 0 \\ 3(x+1), & x > 0 \end{cases}$ For $\lim\limits_{x\to 0}$ f (x) , we have to evaluate $\lim\limits_{x\to 0^-}$ f (x) and $\lim\limits_{x\to 0^+}$ f (x) separately as f(x) is defined differently on left and right of 0. Now, $\lim\limits_{x\to 0^-}$ f (x) = $\lim\limits_{x\to 0^-}$ (2x + 3) = 2 . (0) + 3 = 3 and $\lim\limits_{x\to 0^+}$ f (x) = $\lim\limits_{x\to 0^+}$ 3 (x + 1) = 3 (0 + 1) = 3. Thus we see, $\lim\limits_{x\to 0^-}$ f (x) = $\lim\limits_{x\to 0^+}$ f (x) = 3 ∴ $\lim\limits_{x\to 0}$ f (x) = 3 Now, $\lim\limits_{x\to 1}$ f (x) 3 (x + 1) = 3 × 2 = 6 ∴ $\lim\limits_{x\to 1}$ f (x) = 6 and $\lim\limits_{x\to 0}$ f (x) = 3.