Let $X$ be a Banach space and $A$ is a linear bounded operator on $X$. It is well known that for $|\lambda|> \|A\|,$ we have $$\|(\lambda I - A)^{-1}\| \leq \frac{1}{|\lambda|-\|A\|}.$$
Now, let $f$ be an analytic function at $\infty$ and $\mathcal{C}$ be a circle of radius $r$, such that $r$ is greater than the spectral radius of $A.$ I want to prove that $$\int_{\mathcal{C}}(f(\lambda)-f(\infty))(\lambda I - A)^{-1}d\lambda$$ tends to zero as $r\rightarrow\infty.$ How do it?
Note that $$(\lambda I - A)^{-1} = \dfrac{1}{\lambda} I + \lambda^{-1} A (\lambda I - A)^{-1}$$ The obvious bound $\|\int_C g(\lambda) \; d\lambda\| \le \text{length}(C) \max_C \|g(\lambda)\|$ will do for the second term. For the first, use Cauchy's theorem (either the "outside-the-circle" version, or transform $z = 1/\lambda$ and use the usual version).