= Solution
Set $M_R=\max_{0\le\theta\le\pi}|f(Re^{i\theta})|$. The complex limit at infinity gives $M_R\to0$. Parametrizing the semicircle and using $|e^{i\lambda Re^{i\theta}}|=e^{-\lambda R\sin\theta}$ yields
$$
\left|\int_{\Gamma_R}f(z)e^{i\lambda z}dz\right|\le RM_R\int_0^\pi e^{-\lambda R\sin\theta}\,d\theta.
$$
By symmetry and the supplied sine bound,
$$
\int_0^\pi e^{-\lambda R\sin\theta}d\theta\le2\int_0^{\pi/2}e^{-2\lambda R\theta/\pi}d\theta=\frac\pi{\lambda R}(1-e^{-\lambda R}).
$$
The arc <integral> is therefore at most $(\pi/\lambda)M_R$, tending to zero. This proves \b[the required vanishing], the relevant form of <Jordan's lemma>. No holomorphicity of $f$ is required just for this arc estimate.
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