Question

Question from calculus limit :-

Find :-

Lim x tends to 1

f(x) = ?

Where

f(x) = x^2 - 1 , x≤ 1

= - x^2 - 1 , x>1

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

${\boxed{\boxed{\sf\:{Left\;Hand\;Limit}}}}$

★Assume x = (1 - h)

x → 1

★Hence :-

h → 0

${\boxed{\displaytype \lim_{h\to 0} (f(1-h)}$

${\boxed{\displaytype\lim_{h\to 0}(x^{2}-1)}$

${\boxed{\displaytype\lim_{h\to 0}[(1-h)^{2}-1)]}$

${\boxed{\large{\boxed{\sf\:{Substitute\;the\;value\;of\;h}}}}$

${\boxed{\displaytype\lim_{h\to 0}[(1-0)^{2}-1)]}$

= 1 - 1

= 0 .......... (1)

${\boxed{\boxed{\sf\:{Right\;Hand\;Limit}}}}$

★Assume x = (1 + h)

x → 1

★Hence :-

h → 0

${\boxed{\displaytype\lim_{h\to 0}(-x^{2}-1)}$

${\boxed{\displaytype\lim_{h\to 0}[-(1+h)^{2}-1)}$

= - (1 + 0)² - 1

= -1 - 1

= - 2 .......... (2)

★We get from (1) and (2)

${\boxed{\sf\:{f(1-h)\neq\displaytype\lim_{h\to 0}(f(1-h)}}}$

$\Large{\boxed{\sf\:{Limit\;does\;not\;Exists}}}$

Answer 2

Answer:

f(x) = x^2 - 1 ( x ≤ 1 )

For L. H. L.,

Lim ( x -> 1^ - ) f (x) = [ ( - 1)^2 - 1]

Lim ( x -> 1^ - ) f (x) = 1 - 1 = 0

For R. H. L. ,

Lim ( x -> 1^+) f (x) = - (1) ^2 - 1

Lim ( x -> 1^+) f (x) = - 1 - 1 = - 2

Hence,

L. H.L. ≠ R. H. L.

=> Function is discontinuous.