Suppose I have the following expression
$$\cosh(\rho_{\lambda}(t))-1=\frac{2e^{-2\rho_\lambda(t)}}{1-e^{-2\rho_\lambda(t)}}$$ with $\rho'_{\lambda}(s)=\lambda(1+<\theta(s),f_\lambda(s)>)$ and $\rho_\lambda(0)=0$, where $\theta(s)$ is a path on the unit sphere $\mathbb S^{d-1}\subset \Bbb R^d$ and $f_\lambda(s)$ is a $C^1$-path in $\Bbb R^d$ with $f_\lambda(0)=0$ and $|f'_\lambda|$ is bounded where $|\cdot| $ is the supremum norm.
The claim in the article I am reading is, based on the property that $f_\lambda(0)=0$ and $|f'_\lambda|$ is bounded, the singulartiy of $\frac{2e^{-2\rho_\lambda(t)}}{1-e^{-2\rho_\lambda(t)}}$ at $t=0$ can be killed by multiplication with $f_\lambda(t)$.
I can't understand how the singularity is killed. At time $t=0$, I would have $$\frac{2e^{-2\rho_\lambda(t)}}{1-e^{-2\rho_\lambda(t)}}\bigl\lvert_{t=0}\cdot f_\lambda(0)=0\cdot\infty $$ which is not obvious that this would be a finite number. One could use a argument with L'Hospital and left and right limit. However, the derivative of $\frac{2e^{-2\rho_\lambda(t)}}{1-e^{-2\rho_\lambda(t)}}$ with respect to t will still maintain the term $1-e^{-2\rho_\lambda(t)}$ in the denominator, which causes a singularity at $t=0$. How is the singularity killed in this case? Any advice would be helpful.