We consider the relationship between the minimum degree \(\delta(G\) of a graph and the complexity of recognizing if a graph is \(T\)-tenacious. Let \(T \geq 1\) be a rational number. We first show that if \(\delta(G) \geq \frac{Tn}{T+1}\) then \(G\) is \(T\)-tenacious. On the other hand, for any fixed \(\epsilon > 0\), we show that it is NP-hard to determine if \(G\) is \(T\)-tenacious, even for the class of graphs with \(\delta(G) \geq (\frac{T}{T+1} – \epsilon)n\).