The given function is
When a = 0,
\( \lim\limits_{x \to a^-}f(x)=\lim\limits_{x \to 0^-}(|x|+1)\)
\( \lim\limits_{x \to 0}(-x+1)\)
= 0+1
= 1
Here, it is observed that
\( \lim\limits_{x \to 0^-}f(x)\neq \lim\limits_{x \to 0^+}f(x)\)
\( \lim\limits_{x \to 0}f(x)\) does not exist
When a<0,
\( \lim\limits_{x \to a^-}f(x)=\lim\limits_{x \to a^-}(|x|+1)\)
\( \lim\limits_{x \to a}(-x+1)=-a+1\)
\( \lim\limits_{x \to a^+}f(x)=\lim\limits_{x \to a^+}(|x|+1)\)
\( \lim\limits_{x \to a}(-x+1)\)
-a+1
Thus, limit of f(x) exists at x = a, where a < 0.
When a > 0
\( \lim\limits_{x \to a^-}f(x)=\lim\limits_{x \to a^-}(|x|+1)\)
\( \lim\limits_{x \to a}(-x-1)\)
= a-1
\( \lim\limits_{x \to a^-}f(x)=\lim\limits_{x \to a^+}f(x)=a-1\)
Thus, limit of f(x) exists at x = a, where a > 0.
Thus, \( \lim\limits_{x \to a}f(x)\) exists for all \( a\neq0\)
Answered by Abhisek | 1 year ago