In the soft margin SVM; can someone please give me the intuition of why a high value of the the penalty parameter $C$ causes the SVM to tend towards a hard margin SVM? I am failing to see the logic behind this.
Looking at the objective function of the soft margin SVM: $L_P=\frac{1}{2}||w||^2+C \cdot \Sigma_{i=1}^N{\xi}_i$, I can see that a small value of $C$ (such as one that tends to zero) would mean that we have $L_P=\frac{1}{2}||w||^2+0$, which is the same as the hard margin SVM. Unfortunately this is not the case--a high value of $C$ is what corresponds to a hard margin SVM. Can someone please give me a gentle insight into this concept?