1. **Problem:** Find the value of $a$ such that the function $$f(t) = \begin{cases} \sqrt{4 + t^2} - 2 \, / \, a t^2, & t \neq 0 \\ 0, & t = 0 \end{cases}$$ is continuous everywhere. 2. **Continuity condition:** For $f$ to be continuous at $t=0$, the limit as $t \to 0$ of $f(t)$ must equal $f(0)$. 3. **Evaluate the limit:** $$\lim_{t \to 0} \frac{\sqrt{4 + t^2} - 2}{a t^2}$$ 4. **Simplify the numerator using conjugate:** $$\sqrt{4 + t^2} - 2 = \frac{(\sqrt{4 + t^2} - 2)(\sqrt{4 + t^2} + 2)}{\sqrt{4 + t^2} + 2} = \frac{4 + t^2 - 4}{\sqrt{4 + t^2} + 2} = \frac{t^2}{\sqrt{4 + t^2} + 2}$$ 5. **Rewrite the limit:** $$\lim_{t \to 0} \frac{t^2}{a t^2 (\sqrt{4 + t^2} + 2)} = \lim_{t \to 0} \frac{1}{a (\sqrt{4 + t^2} + 2)}$$ 6. **Evaluate the limit as $t \to 0$:** $$\frac{1}{a (\sqrt{4 + 0} + 2)} = \frac{1}{a (2 + 2)} = \frac{1}{4a}$$ 7. **Set the limit equal to $f(0)$ for continuity:** $$\frac{1}{4a} = 0$$ 8. **Solve for $a$:** This equation has no solution for finite $a$ because $\frac{1}{4a} = 0$ implies $a \to \infty$. 9. **Re-examine $f(0)$:** The problem states $f(0) = 0$. 10. **Check if the original function is correctly interpreted:** The problem likely means $$f(t) = \frac{\sqrt{4 + t^2} - 2}{a t^2} \text{ for } t \neq 0, \quad f(0) = 0$$ 11. **For continuity, limit must equal 0:** $$\lim_{t \to 0} \frac{\sqrt{4 + t^2} - 2}{a t^2} = 0$$ 12. **From step 6, limit is $\frac{1}{4a}$, so:** $$\frac{1}{4a} = 0 \implies \text{No finite } a \text{ satisfies this}$$ 13. **Conclusion:** The function cannot be continuous at $t=0$ for any finite $a$ if $f(0) = 0$. 14. **Alternative interpretation:** If $f(0) = L$ and we want continuity, then $$\lim_{t \to 0} f(t) = f(0) = L$$ 15. **If $f(0) = 0$, then $\frac{1}{4a} = 0$ no solution. If $f(0) = \frac{1}{4a}$, then $a$ can be any real number except zero.** 16. **Therefore, to make $f$ continuous everywhere, define $f(0) = \lim_{t \to 0} f(t) = \frac{1}{4a}$.** 17. **If the problem requires $f(0) = 0$, then no finite $a$ makes $f$ continuous everywhere.** **Final answer:** $$\boxed{a = \frac{1}{4 f(0)}}$$ If $f(0) = 0$, no finite $a$ exists.