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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 24 of 72
Marks: +1, -0
Find lim⁡x→1\lim\limits_{x\rightarrow 1} f (x) , where f (x) = {x2−1x≤1−x2−1x>1\begin{cases} x^2-1 & x \le 1 \\ -x^2-1 & x > 1 \end{cases}
Solution:  
We have, f (x) = {x2−1x≤1−x2−1x>1\begin{cases} x^2-1 & x \le 1 \\ -x^2-1 & x > 1 \end{cases}
Now, lim⁡x→1\lim\limits_{x\rightarrow 1} f (x) = lim⁡x→1\lim\limits_{x\rightarrow 1} (x2−1)(x^2-1) = 121^2 - 1 = 0.
and lim⁡x→1+\lim\limits_{x\rightarrow 1^{+}} f (x) = lim⁡x→1+\lim\limits_{x\rightarrow 1^{+}} (−x2−1)(-x^2-1) = −(1)2-(1)^2 - 1
= - 1 - 1 = - 2.
Since lim⁡x→1−\lim\limits_{x\rightarrow 1^{-}} f (x) ≠ lim⁡x→1+\lim\limits_{x\rightarrow 1^{+}} f (x)
∴ lim⁡x→1\lim\limits_{x\rightarrow 1} f (x) does not exist.
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