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Exercise 9.1 · Q23

Q.Verify the existence of lim⁡x→1f(x)\displaystyle\lim_{x\to1}f(x), where
[!FORMULA] f(x)={∣x−1∣x−1,x≠10,x=1f(x)=\begin{cases}\dfrac{|x-1|}{x-1}, & x\ne1\\ 0, & x=1\end{cases}

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Step 1. Simplify for x<1x<1. Here x−1<0x-1<0, so ∣x−1∣=−(x−1)|x-1|=-(x-1), giving f(x)=−(x−1)x−1=−1f(x)=\dfrac{-(x-1)}{x-1}=-1.

Step 2. Simplify for x>1x>1. Here x−1>0x-1>0, so ∣x−1∣=x−1|x-1|=x-1, giving f(x)=x−1x−1=1f(x)=\dfrac{x-1}{x-1}=1.

Step 3. Compute the one-sided limits. lim⁡x→1−f(x)=−1\displaystyle\lim_{x\to1^-}f(x)=-1 and lim⁡x→1+f(x)=1\displaystyle\lim_{x\to1^+}f(x)=1.

Step 4. Compare. Since −1≠1-1\ne1, the two one-sided limits disagree, so lim⁡x→1f(x)\displaystyle\lim_{x\to1}f(x) does not exist. …

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