Prove that a regularized Newton-type step $x_{k+1} = x_{k} - (B_{k} + \alpha I)^{-1} \nabla f(x_{k})$ with $B_{k}$ a Hessian approximation, $\alpha$ a positive scalar and $I$ the identity matrix converges to a small gradient step $x_{k+1} = x_{k} - \frac{1}{\alpha} \nabla f(x_{k}) $ as $ \alpha \rightarrow \infty$.
I am not sure how to approach this problem. I am thinking to multiple the equation out so I have $x_{k+1} = x_{k} - B_{k}^{-1} \nabla f(x) - \alpha I^{-1} \nabla f(x_{k})$ so the first part $x_{k+1} = x_{k} - B_{k}^{-1} \nabla f(x)$ is not a regularized problem.
Is this a correct first step? Then how should I approach it?
First, the question as stated is problematic since the expression to which you converge contains $\alpha$. (What does it mean for something to converge to $\frac{1}{\alpha} I$ if $\alpha$ is tending to infinity?) But there is some sense that the two expressions get "closer" as $\alpha \to \infty$.
Since $B_k$ is positive semi-definite, has nonnegative eigenvalues $\lambda_1,\ldots,\lambda_n$ with respect to an orthonormal eigenbasis. Then $(B_k + \alpha I)^{-1}$ has eigenvalues $\frac{1}{\lambda_1 + \alpha}, \ldots, \frac{1}{\lambda_n + \alpha}$ with respect to the same eigenbasis. You want to show that this is close to $\frac{1}{\alpha} I$, i.e. the matrix with eigenvalues $\frac{1}{\alpha},\ldots, \frac{1}{\alpha}$ (with respect to the same eigenvectors) when $\alpha$ tends to infinity. For instance, the ratio of the respective eigenvalues is $\frac{\alpha}{\lambda_i+\alpha}$ which tends to $1$ as $\alpha \to \infty$. Also, the difference in eigenvalues is $\frac{\lambda_i}{\alpha(\lambda_i+\alpha)}$ which tends to zero as $\alpha \to \infty$. You should clarify the problem statement to figure out what exactly you need to show here.