= Solution
{c}
Write $c=c_1=c_3$, so the friction <memory kernel> is $\kappa_c(t)=ce^{-ct}$ and has area one for every $c>0$. Its <Laplace transform> is $\widetilde\kappa_c(s)=c/(s+c)$. The <generalized fluctuation-dissipation relation> therefore gives
$$
\widetilde\chi_c(s)=\frac5{s+c/(s+c)}
=\frac{5(s+c)}{s^2+cs+c}
\longrightarrow\frac5{s+1}
\quad(c\to\infty).
$$
To verify the pointwise time-domain limit, rather than relying only on <Laplace transform> convergence, consider $c>4$. The two roots
$$
\lambda_\pm=\frac{-c\pm\sqrt{c^2-4c}}2
$$
are negative and satisfy $\lambda_+\to-1$, $\lambda_-\to-\infty$. Since $\chi_c(0)=5$ and $\chi_c'(0)=0$, the inverse <Laplace transform> is
$$
\chi_c(\tau)=5\frac{-\lambda_-e^{\lambda_+\tau}+\lambda_+e^{\lambda_-\tau}}{\lambda_+-\lambda_-}.
$$
The coefficient of the slow exponential tends to $5$, and that of the fast exponential tends to zero. Hence
$$
\boxed{\lim_{c_1\to\infty}\chi(\tau)=5e^{-\tau},\qquad \tau\geq0.}
$$
This is the <velocity> <autocorrelation function> of an <Ornstein-Uhlenbeck process> satisfying $dV=-V\,dt+\sqrt{10}\,dW$. The friction <memory kernel> becomes local in time. Consistently, the even noise <autocorrelation> $5ce^{-c|\tau|}$ tends to $10\delta(\tau)$ in the sense of <distributions>, fixing the <white noise> normalization. The finite-$c$ right derivative $\chi_c'(0)=0$ does not converge to the right derivative $-5$ of the limiting function: there is a short initial relaxation layer.
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