Behaviour being used: Adaptive mode
Minimum fraction: 0
Question summary: Honako funtzioa kontuan hartuz: [f left parenthesis x right parenthesis space equals space fraction numerator # p o l i 1 over denominator # p o l i 2 end fraction] Bere definizio eremua hau da: [straight real numbers space minus space open curly brackets # d comma # e close curly brackets] Asintota bertikaletako alboko limiteak kalkulatu: [stack L i m with x rightwards arrow # d to the power of minus below space f left parenthesis x right parenthesis] = {infinito_negativo; oo_} [stack L i m with x rightwards arrow # d to the power of plus below space f left parenthesis x right parenthesis] = {oo_; infinito_negativo} [stack L i m with x rightwards arrow # e to the power of minus below space f left parenthesis x right parenthesis] = {infinito_negativo; oo_} [stack L i m with x rightwards arrow # e to the power of plus below space f left parenthesis x right parenthesis] = {infinito_negativo; oo_} Eta asintota horizontala dauka y = _____ baliorako. OHARRAK: Emaitza zatikia bada, zatiki bezala idatzi (adibidez 3/4)
Right answer summary: part 1: oo_; part 2: infinito_negativo; part 3: infinito_negativo; part 4: oo_; part 5: 2
Question state: todo