Holef(x) = (x² − 1) / (x − 1)
f(x)L⁻ and L⁺x = capproach
From the leftx → c⁻
x =0.5
f(x) =1.5
↓ as ε → 0
L⁻ =2
From the rightx → c⁺
x =1.5
f(x) =2.5
↓ as ε → 0
L⁺ =2
Limit at x = cc = 1
Left limit L⁻
2
lim x→c⁻ f(x)
Right limit L⁺
2
lim x→c⁺ f(x)
Limit exists. L⁻ and L⁺ agree at 2, so lim x→1 f(x) = 2.limit exists
f(c) = undefined. The limit exists but f isn’t defined at c. Set f(c) = 2 and the function becomes continuous.removable
Appliedc=1ε=0.5L⁻ = L⁺ ⇒ limit exists