Is it (for some reason) true that
$\lim\_{c\to 0^+}\int\_c^{\pi/2}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2}$? 


**Numerical evidence (from Mathematica):**


* when $c=1/5$, the integral is $\approx 1.578$.
* when $c=1/10$, the integral is $\approx 1.575$.
* when $c=1/100$, the integral is $\approx 1.571$.


**Geometric motivation:**  

By Clairaut's relation, if for some $T$,
$$\int\_c^{T}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2},$$
then there are two three distinct geodesics between the points $(T,0, \tfrac{1}{2}T^2)$ and $(-T,0, \tfrac{1}{2}T^2)$ on the paraboloid of revolution $z=\tfrac{1}{2}r^2$: one through the point $(0,0,0)$, $(0, c, \tfrac{1}{2}c^2)$ and one through the point $(0, -c, \tfrac{1}{2}c^2)$.


**Edit:** Deleted the comment about injectivity radius which concluded my original question -- it was false. Note that the answers of Nemo and Christian Remling, below, appear to work for **any** integration bound $T$, not just for $T=\pi/2$: for any $T>0$,
$$\lim\_{c\to 0^+}\int\_c^{T}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2}.$$



Identity involving an improper integral (with geometric application)
Since the main contribution to the integral comes from $t<<1$, analytically one has
\begin{align}
\lim\_{c\to 0^+}\int\_c^{\pi/2}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt&=\lim\_{c\to 0^+}\int\_c^{\pi/2}\frac{c}{t}\sqrt\frac{1}{t^2-c^2}dt\\
&=\lim\_{c\to 0^+}\int\_{2/\pi}^{1/c}c\sqrt\frac{1}{1-c^2t^2}dt\\
&=\lim\_{c\to 0^+}\int\_{2c/\pi}^{1}\sqrt\frac{1}{1-t^2}dt\\
&=\int\_{0}^{1}\sqrt\frac{1}{1-t^2}dt\\
&=\frac\pi{2}
\end{align}


