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+# Gegeben
+* Anzeigefrequenz: $f_\text{Anzeige} = \qty{14100}{\kilo\hertz}$
+* Genauigkeit: $\pm \qty{0,00001}{\percent}$
+
+# Lösungsweg
+$\qty{1}{\percent}$ entspricht $1 \cdot {10^{-2}}$;  
+$\qty{0,00001}{\percent}$ entspricht $0,1 \cdot {10^{-6}}$.  
+
+$\Delta f = \qty{14100}{\kilo\hertz} \cdot \qty{0,00001}{\percent} = \qty{14,1 \cdot \cancel{10^6}}{\hertz} \cdot 0,1\cdot \cancel{10^{-6}} = \qty{1,41}{\hertz}$
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