If $$f(x)=\left\{\begin{matrix} 2x+3, & x\leq 0\\ 3(x+1), & x > 0\end{matrix}\right.$$. Find $$\displaystyle\lim_{x\rightarrow 0}f(x)$$ and $$\displaystyle\lim_{x\rightarrow 1}f(x)$$.
We have,
$$f(x)=\left\{\begin{matrix} 2x+3, & x \leq 0\\ 3(x+1) & x \geq 0\end{matrix}\right.$$
For the first case:
$$L.H.L:\ \ \displaystyle\lim_{x\rightarrow 0^-}f(x)=\displaystyle\lim_{x\rightarrow 0^-}(2x+3)=2\times 0+3=3$$
$$R.H.L:\ \ \displaystyle\lim_{x\rightarrow 0^+}f(x)=\displaystyle\lim_{x\rightarrow 0^+}3(x+1)=3(0+1)=3$$
So, $$\displaystyle\lim_{x\rightarrow 0}f(x)$$ exists and is equal to $$3$$
For the second case:
$$L.H.L:\ \ \displaystyle\lim_{x\rightarrow 1^-}f(x)=\displaystyle\lim_{x\rightarrow 1^-}2x+3=2\times 1+3=5$$
$$R.H.L:\ \ \displaystyle\lim_{x\rightarrow 1^+}f(x)=\displaystyle\lim_{x\rightarrow 1^+}3(x+1)=3(1+1)=6$$
Since, $$\displaystyle\lim_{x\rightarrow 1^-}f(x)\neq \displaystyle\lim_{x\rightarrow 1^+}f(x)$$
Hence, $$\displaystyle\lim_{x\rightarrow 1^-}f(x)$$ does not exist.