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Question 120 of 144

Q.For the function f(x)={x+2,x>0x−2,x<0f(x)=\begin{cases}x+2, & x>0\\ x-2, & x<0\end{cases}

(a) lim⁡x→2−f(x)=−1\lim\limits_{x\to 2^-} f(x)=-1
(b) lim⁡x→0f(x)\lim\limits_{x\to 0} f(x) does not exist
(c) lim⁡x→0−f(x)=−1\lim\limits_{x\to 0^-} f(x)=-1
(d) lim⁡x→0+f(x)=1\lim\limits_{x\to 0^+} f(x)=1
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2019MCQ· 1mImportance★★★★★
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As x→0+x\to 0^+, f(x)=x+2→2f(x)=x+2\to 2; as x→0−x\to 0^-, f(x)=x−2→−2f(x)=x-2\to -2. Since these one-sided limits differ, the overall limit at x=0x=0 does not exist, matching option (b); the other three stated values are each incorrect when checked directly.

Check (a): lim⁡x→2−f(x)\lim_{x\to 2^-} f(x) — near x=2x=2 we're on the x>0x>0 branch, so f(x)=x+2→2+2=4f(x)=x+2\to 2+2=4, not −1-1. False.

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