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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Find $ \lim\limits_{ x \to 1} f(x)$, where $f(x)=\left\{\begin{array}{1 1}x^2-1, &x \leq 1\\-x^2-1, &x>1\end{array}\right.$

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The given function is
$f(x)=\left\{\begin{array}{1 1}x^2-1, &x \leq 1\\-x^2-1, &x>1\end{array}\right.$
$ \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1} [ x^2-1] = 1^2-1 = 1-1 = 0$
$ \lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1} [ -x^2-1] = -1^2-1 = -1-1 = -2$
It is observed that $ \lim\limits_{x \to 1^-}f(x) \neq \lim\limits_{x \to 1^+}f(x).$
Hence, $\lim\limits_{x \to 1}f(x)$ does not exist.
answered Apr 5, 2014 by thanvigandhi_1
 

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