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Question

Find $$\displaystyle \lim_{x\rightarrow 1}f\left ( x \right ),$$ where $$f(x) =$$ $$\displaystyle \begin{cases}x^{2} -1,& \text{ } x\leq 1 \\-x^{2}-1, & \text{ } x> 1\ \, \end{cases}$$


Solution

$$\mathrm{f}(x) = \displaystyle \begin{cases}x^{2} -1,& \text{ } x\leq 1 \\-x^{2}-1, & \text{ } x> 1\ \, \end{cases}$$
L.H.L $$=\displaystyle \lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1}\left [ x^{2}-1 \right ]=1^{2}-1=1-1=0$$
R.H.L $$=\displaystyle \lim_{x\rightarrow 1^{+}}f\left ( x \right )=\lim_{x\rightarrow 1}\left [ -x^{2}-1 \right ]=-1^{2}-1=-1-1=-2$$
Clearly L.H.L $$\neq $$ R.H.L
Hence $$\displaystyle \lim_{x\rightarrow 1}f\left ( x \right )$$ does not exist.

Mathematics
NCERT
Standard XI

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