f/g = x sin(1/x) → 0, but f′/g′ = 2x sin(1/x) − cos(1/x) has no limit. L'Hôpital goes one way only.
Counterexample bank
Part C is won by knowing which tempting claims are false. 148 counterexamples; 85 free. The rest come with the Notes pack.
AllODEPDEabeliananalyticityanovabairecalculus of variationscanonical formscauchy-riemanncompactnesscompletenesscomplexconformalconnectednesscontinuitycontour integrationconvergencedifferentiationdistributionsfieldsfixed pointgaloisgreengroupsharmonicinjectivityintegral equationsintegrationlaurentlimsuplinear algebraliouvillelpmarkovmeasuremechanicsmetric spacesmultivariablemultivariatenumericalp-groupspermutationspolynomialspower seriesprobabilityquadratic formsreal analysisregressionresiduesringssamplingseparabilityseparationsequencesseriessingularitiesstabilitystatisticssylowtestingtopologyuniform convergencewronskianzeros
#1 · Analysis & Linear Algebra › Continuity and Differentiation
“Differentiable ⇒ continuously differentiable” — false
Counterexample locked — unlock with Notes + PYQ
differentiation
#2 · Analysis & Linear Algebra › Continuity and Differentiation
“If lim f/g exists (0/0 form) then lim f′/g′ exists” — false
Counterexample: f(x) = x² sin(1/x), g(x) = x as x → 0
differentiation
#3 · Analysis & Linear Algebra › Continuity and Differentiation
“f′(x₀) = 0 and f′ changes sign nowhere near x₀ ⇒ f is constant near x₀” — false
Counterexample locked — unlock with Notes + PYQ
differentiation