I have encountered the following trick that people in C*-algebra use; but frankly I don't understand why really this is true.
Let $A$ be a unital C*-algebra acting non-degeneratly on a Hilbert space $H$ via its universal representation. Take a positive element $h\in A$, $\|h\|\leqslant 1$. How can one prove that the identity operator $I_H$ on $H$ can be written as
$$I_H=\mbox{SOT}\!-\!\lim_{n\to \infty} (h+\tfrac{1}{n}I_H)^{-1} h?$$
In general it is not the identity of $H$ what you get, but the range projection of $h$. For example if $$ h=\begin{bmatrix}1/2&0\\0&0\end{bmatrix}, $$ then $$ (h+1/n I)^{-1}h=\begin{bmatrix}\frac1{1/2+1/n}&0\\0&n\end{bmatrix}\,\begin{bmatrix}1/2&0\\0&0\end{bmatrix}=\begin{bmatrix}\frac{1/2}{1/2+1/n}&0\\0&0\end{bmatrix}\to\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ To show that this is always the case, take $h$ positive. Then, for $\xi=h\eta$, $$ (h+\frac1n\,I)^{-1}h\,\xi-\xi=(h+\frac1n\,I)^{-1}\left[h\,\xi-(h+\frac1n\,I)\xi\right]=(h+\frac1n\,I)^{-1}\left[-\frac1n\,\xi\right]\\ =-\frac1n\,(h+\frac1n\,I)^{-1}\,h\eta. $$ As $\|(h+\frac1n\,I)^{-1}h\|\leq1$ (just because $|t/(t+1/n)|\leq1$ for $t\geq0$), we have $$ \|(h+\frac1n\,I)^{-1}h\,\xi-\xi\|=\frac1n\,\|(h+\frac1n\,I)^{-1}\,h\eta\|\leq\frac{\|\eta\|}n\to0, $$ and the result extends to the closure of the range of $h$.
Finally, for $\xi\in(\text{ran}\,h)^\perp$, $h\xi=0$, so $(h+\frac1n\,I)^{-1}h\xi=0$.