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  • If f x equals mode x plus 1 for x lesser tahn 0 . For what value (s) of a does lim x tends to a f (x) exists?

30. If  f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.  For what value (s) of a does \lim_{x \rightarrow a } f (x) exists ? 

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f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.

Limit at x = a exists when the right-hand limit is equal to the left-hand limit. So,

Case 1: when a = 0

The right-hand Limit or  Limit at x=0^+

\lim_{x \rightarrow 0^+} f (x) = \lim_{x \rightarrow 0^+} |x|-1=1-1=0

The left-hand limit or Limit at x=0^-

\lim_{x \rightarrow 0^-} f (x) = \lim_{x \rightarrow 0^-} |x|+1=0+1=1

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

Case 2: When a < 0 

The right-hand Limit or  Limit at x=a^+

\lim_{x \rightarrow a^+} f (x) = \lim_{x \rightarrow a^+} |x|-1=a-1

The left-hand limit or Limit at x=a^-

\lim_{x \rightarrow a^-} f (x) = \lim_{x \rightarrow a^-} |x|-1=a-1

Since LHL = RHL, Limit exists at x = a and is equal to a-1.

Case 3: When a > 0 

The right-hand Limit or  Limit at x=a^+

\lim_{x \rightarrow a^+} f (x) = \lim_{x \rightarrow a^+} |x|+1=a+1

The left-hand limit or Limit at x=a^-

\lim_{x \rightarrow a^-} f (x) = \lim_{x \rightarrow a^-} |x|+1=a+1

Since LHL = RHL, Limit exists at x = a and is equal to a+1

Hence,

The limit exists at all points except at x=0.

Posted by

Pankaj Sanodiya

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