have the right to i ask because that the limit as \$x\$ approaches \$lim_x o 0-ln x\$? you re welcome explain. Its because since the border of a function only exist if the lim together \$x\$ viewpoints some number \$n\$ indigenous both the optimistic and an unfavorable side is the same, ns not sure that im convinced that the limit as \$x\$ viewpoints \$0\$ because that \$ln x\$ exists. I know its an adverse infinity indigenous the optimistic side, but from the an adverse side?

The domain the \$ln(x)\$ is only positive reals, so the left-hand limit at 0 doesn"t really do sense.

You are watching: Limit of ln x as x approaches 0

For a limit to exist, the two limits approaching from the left and right side require to enhance up.

I.e. We require \$\$limlimits_x o0^-ln(x) = limlimits_x o0^+ln(x)\$\$ to it is in true.

Since \$ln(x)\$ is not identified for \$xleq 0\$ assuming us are evaluating over the reals, the left-hand border can"t be evaluated, and thus the limit does no exist.

If friend are analyzing over the facility numbers, that"s a somewhat various story, however given her wording, I"ve assumed that we"re talking around the reals here.

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