Find _ x 1 f(x) where f(x) = * 20 c x^2 - 1, & x 0 - x^2 - 1, & x… - Vidyadip

Question

Find $\mathop {\lim }\limits_{x \to 1} f(x)$ where f(x) = $\left\{ {\begin{array}{*{20}{c}} {{x^2} - 1,}&{x \le 0} \\ { - {x^2} - 1,}&{x > 1} \end{array}} \right.$

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Answer

Here $\mathop {\lim }\limits_{x \to 1} f(x)$ = $\left\{ {\begin{array}{*{20}{c}} {{x^2} - 1,}&{x \le 0} \\ { - {x^2} - 1,}&{x > 1} \end{array}} \right.$
L.H.L. $\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} ({x^2} - 1)$
Put x = 1 - h as $x \to 1,\;h \to 0$
$\therefore \;\mathop {\lim }\limits_{h \to 0} [{(1 + h)^2} - 1] = \mathop {\lim }\limits_{h \to 0} [1 + {h^2} - 1]$
R.H.L. $= \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} ( - {x^2} - 1)$
Put x = 1 + h as $x \to 1,\;h \to 0$
$\therefore \;\mathop {\lim }\limits_{h \to 0} [ - {(1 + h)^2} - 1] = \mathop {\lim }\limits_{h \to 0} [ - 1 - {h^2} - 2h - 1]$
$= - {(0)^2} - 2 \times 0 - 2 = - 2$
$\therefore$ LHL $\neq$ RHL
Therefore limit of given function does not exist

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