$\underset{_{a\rightarrow 0}}{\lim }L_{a}^{\ast }\left( L_{a}L_{a}^{\ast }\right) ^{-1}L_{a}=?$

Let $$\varepsilon >0$$ and let $$L$$ be a bounded operator acting on a Hibert space such that $$L_{\lambda }=L-\lambda I$$ is surjective of every $$\lambda$$ such that $$\varepsilon >\left\vert \lambda \right\vert >0$$.

I'm trying, by applying the Cauchy criterion, to show that the limit

$$\underset{_{\lambda \rightarrow 0}}{\lim }L_{\lambda }^{\ast }\left( L_{\lambda }L_{\lambda }^{\ast }\right) ^{-1}L_{\lambda }$$

exists, where $$L_{\lambda }^{\ast }=L^{\ast }-\overline{\lambda }I.$$ If this is not true, can you give me some counter-example ?

Thank you !

Setting $$L=0$$ shows $$L_\lambda = -\lambda I$$, and the operator $$L_\lambda^*(L_\lambda L^*_\lambda)^{-1}L_\lambda$$ is equal to the identity. So it seems to disprove convergence is going to be hard.
It is quite easy to show that $$L_\lambda^*(L_\lambda L^*_\lambda)^{-1}L_\lambda =I$$ holds for finite-dimensional $$H$$ or normal $$L$$.
Here is a partial answer for the general case: Using singular value decomposition of $$L=UT_fV^*$$, where $$U$$ is partial isometry ($$U:N(U)^\perp\to R(U)$$ is isometry), $$V$$ isometry, $$T_f$$ multiplication operator on some $$L^2(\mu)$$.
Then $$L_\lambda^*(L_\lambda L^*_\lambda)^{-1}L_\lambda = VT_{\bar f-\bar\lambda}U^* (UT_{f-\lambda}V^* VT_{\bar f-\bar\lambda}U^*)^{-1}UT_{f-\lambda}V^* =VT_{\bar f-\bar\lambda}U^* (UT_{|f|^2+|\lambda|^2}U^*)^{-1}UT_{f-\lambda}V^*.$$ If $$U$$ would be a (full) isometry, then clearly this operator reduces to the identity.