If f(x) = $$\{|x|+1,x<0,0,x=0,|x|-1,x>1$$ For what value (s) of does $$\lim\limits_{x \to a}f(x)$$exists?

Asked by Pragya Singh | 1 year ago |  58

##### Solution :-

The given function is

When a = 0,

$$\lim\limits_{x \to a^-}f(x)=\lim\limits_{x \to 0^-}(|x|+1)$$

$$\lim\limits_{x \to 0}(-x+1)$$

= 0+1

= 1

Here, it is observed that

$$\lim\limits_{x \to 0^-}f(x)\neq \lim\limits_{x \to 0^+}f(x)$$

$$\lim\limits_{x \to 0}f(x)$$ does not exist

When a<0,

$$\lim\limits_{x \to a^-}f(x)=\lim\limits_{x \to a^-}(|x|+1)$$

$$\lim\limits_{x \to a}(-x+1)=-a+1$$

$$\lim\limits_{x \to a^+}f(x)=\lim\limits_{x \to a^+}(|x|+1)$$

$$\lim\limits_{x \to a}(-x+1)$$

-a+1

Thus, limit of f(x) exists at x = a, where a < 0.

When a > 0

$$\lim\limits_{x \to a^-}f(x)=\lim\limits_{x \to a^-}(|x|+1)$$

$$\lim\limits_{x \to a}(-x-1)$$

= a-1

$$\lim\limits_{x \to a^-}f(x)=\lim\limits_{x \to a^+}f(x)=a-1$$

Thus, limit of f(x) exists at x = a, where a > 0.

Thus, $$\lim\limits_{x \to a}f(x)$$ exists for all $$a\neq0$$

Answered by Abhisek | 1 year ago

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