Find lim f(x),x∈1 f(x) = {x^2-1,x≤1 -x^2-1,x>1.

Question

Find \(\lim\limits_{x \to 1}\) f(x), if f(x) = \(\begin{cases} x^2-1 , x \leq{1}\\ -x^2-1, x>{ 1} \end{cases} \).

lim f(x),x∈1

f(x) = {x²-1,x≤1 -x²-1,x>1.

Answer 1

Given,

f(x) = {x²-1,x≤1 -x²-1,x>1.

To find : lim f(x),x∈1

To limit to exist we know,

\(\lim\limits_{x \to h^+}\)f(x) = \(\lim\limits_{x \to h^-}\)f(x) = \(\lim\limits_{x \to h}\)f(x) …….(1)

Thus,

To find the limit using the concept :

\(\lim\limits_{x \to 1^+}\)f(x) = \(\lim\limits_{x \to 1^-}\)f(x) = \(\lim\limits_{x \to 1}\) f(x) ……(2)

From above equations,

 \(\lim\limits_{x \to 1^+}\)f(x) ≠ \(\lim\limits_{x \to 1^-}\)f(x)

Thus,

The limit \(\lim\limits_{x \to 1}\)f(x) does not exists.