L.H.L,
= \(\lim\limits_{x \to 2^-}f(x)
\)
= \(\lim\limits_{x \to 2^-}(2x+3)
\)
= 2 + 5
= 7
And R.H.L,
= \(\lim\limits_{x \to 2^+}f(x)\lim\limits_{x \to 2^+}(x+5)\)
= 2 + 5
= 7
Since \(\lim\limits_{x \to 2^-}f(x)
\) = \(\lim\limits_{x \to 2^+}f(x)
\)=7
∴ \(\lim\limits_{x \to 2^+}f(x)
\) exists and is equal to 7.