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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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$f(x)=\left\{\begin{array}{1 1}|x|+1,&x<0\\0,&x=0 \\|x|-1,&x>0\end{array}\right.$ \[\] For what value(s) of a does $ \lim\limits_{x \to a}f(x)$ exists?

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The given function is
$f(x)=\left\{\begin{array}{1 1}|x|+1,&x<0\\0,&x=0 \\|x|-1,&x>0\end{array}\right.$
When $a = 0$
$ \lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-}(|x|+1)$
$ = \lim\limits_{x \to 0}(-x+1) \quad \quad [ If\: x < 0, \: |x|=-x]$
$ = -0+1$
$ = 1$
$ \lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+}(|x|-1)$
$ = \lim\limits_{x \to 0}(x-1) \quad \quad [ If\: x > 0, \: |x|=x]$
$ = 0-1$
$ = -1$
Here it is observed that $ \lim\limits_{x \to 0^-} f(x) \neq \lim\limits_{x \to 0^+} f(x)$
$ \therefore \lim\limits_{x \to 0} f(x)$ does not exist.
when $ a < 0$
$ \lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^-}(|x|+1)$
$ = \lim\limits_{x \to a}(-x+1) \quad \quad [ If\: x < a < 0 \Rightarrow |x|=-x]$
$ = -a + 1$
$ \lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+}(|x|+1)$
$ = \lim\limits_{x \to a}(-x+1) \quad \quad [ If\: a < x < 0 \Rightarrow |x|=-x]$
$ = -a + 1$
$ \therefore \lim\limits_{x \to a^-}f(x) = \lim\limits_{x \to a^+}f(x) = -a+1$
Thus, the limit of $f(x)$ exists at $x = a$, where $a < 0$
When $ a > 0$
$ \lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^-}(|x|-1)$
$ = \lim\limits_{x \to a} (x-1) \qquad [0 < x < a \Rightarrow |x| = x]$
$ = a-1$
$ \lim\limits_{x \to a^+} f(x) = \lim\limits_{x \to a^+}(|x|-1)$
$ = \lim\limits_{x \to a} (x-1) \qquad [0 < a < x \Rightarrow |x| = x]$
$ = a-1$
$ \therefore \lim\limits_{x \to a^-}f(x) = \lim\limits_{x \to a^+}f(x) = a-1$
Thus, the limit of $f(x)$ exists at $x = a$, where $a > 0$
Thus, $ \lim\limits_{x \to a}f(x)$ exist for all $a \neq 0$
answered Apr 7, 2014 by thanvigandhi_1
 

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