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If f(x)={{:(,1,(x lt 0)),(,2x+1,(x le 1)...

If `f(x)={{:(,1,(x lt 0)),(,2x+1,(x le 1)),(,3x,(x gt 1)):}" then "underset(x to 1)"Lt" f(x)=`

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The correct Answer is:
`Lt_(x to 1^(-))f(x)=3, Lt_(x to1^(+)) f(x)=3`, limits exists.
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