SVM behavior when regularization parameter equals 0 I read on this Wikipedia page the following about soft-margin SVM:




  "The parameter $λ$ determines the trade-off between increasing the
  margin size and ensuring that the $x_i$ lie on the correct side of the
  margin. Thus, for sufficiently small values of $λ$, the second term in
  the loss function will become negligible, hence, it will behave
  similar to the hard-margin SVM, if the input data are linearly
  classifiable, but will still learn if a classification rule is viable
  or not."


I can't understand why in the case that λ=0 the algorithm will behave like hard-margin SVM. If λ=0, it seems to me that the algorithm won't have any reason to perform any optimization on the margin. Doesn't it just become a perceptron in that case, since the algorithm only "cares" about classifying all the train data correctly, while not reaching any optimal solution regarding the margin?

I'll appreciate a clarification about the issue, please.
First, the remark from the wikipedia article deals with small (positive) values of $\lambda$, not $\lambda=0$.  Indeed, if $\lambda=0$, then every separating hyperplane achieves the minimum score of 0.

If the data is linearly separable, then taking $\lambda$ small enough that the first term dominates ensures that minimizing the loss will require taking a separating hyperplane to zero out the first term, and subject to that minimizing the second term is equivalent to the original hard SVM.