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Exercises · Q9

Q.Examine the continuity of f(x)={x+2,x<13,x=1x2+1,x>1f(x)=\begin{cases}x+2, & x<1\\ 3, & x=1\\ x^{2}+1, & x>1\end{cases} at x=1x=1.

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Step 1 — Left-hand limit.

lim⁡x→1−f(x)=lim⁡x→1−(x+2)=1+2=3\lim_{x\to1^{-}}f(x)=\lim_{x\to1^{-}}(x+2)=1+2=3

Step 2 — Right-hand limit.

lim⁡x→1+f(x)=lim⁡x→1+(x2+1)=12+1=2\lim_{x\to1^{+}}f(x)=\lim_{x\to1^{+}}(x^{2}+1)=1^{2}+1=2

Step 3 — Value of the function at the point.

f(1)=3(given directly by the middle branch)f(1)=3 \quad\text{(given directly by the middle branch)}

Step 4 — Apply the continuity test. LHL (=3)≠(=3)\neq RHL (=2)(=2), so lim⁡x→1f(x)\displaystyle\lim_{x\to1}f(x) does not exist at all — meaning condition 2 of continuity fails regardless of what f(1)f(1) equals. ff is therefore discontinuous at x=1x=1 (specifically, a jump discontinuity). …

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