1. **State the problem:** We want to find the limit $$\lim_{x \to \infty} \frac{\int_0^{2x} \sqrt{1+t^2} \, dt}{x^2}.$$\n\n2. **Recall the integral and limit concepts:** The integral $$\int_0^{2x} \sqrt{1+t^2} \, dt$$ represents the area under the curve $$y=\sqrt{1+t^2}$$ from 0 to $$2x$$. We want to understand how this area grows compared to $$x^2$$ as $$x$$ becomes very large.\n\n3. **Find an antiderivative:** To evaluate the integral, use the formula for $$\int \sqrt{1+t^2} \, dt$$: $$\int \sqrt{1+t^2} \, dt = \frac{t}{2} \sqrt{1+t^2} + \frac{1}{2} \ln\left(t + \sqrt{1+t^2}\right) + C.$$\n\n4. **Evaluate the definite integral:** Substitute the limits 0 and $$2x$$: $$\int_0^{2x} \sqrt{1+t^2} \, dt = \left[ \frac{t}{2} \sqrt{1+t^2} + \frac{1}{2} \ln\left(t + \sqrt{1+t^2}\right) \right]_0^{2x}.$$\n\n5. **Calculate the expression at the bounds:**\nAt $$t=2x$$: $$\frac{2x}{2} \sqrt{1+(2x)^2} + \frac{1}{2} \ln\left(2x + \sqrt{1+(2x)^2}\right) = x \sqrt{1+4x^2} + \frac{1}{2} \ln\left(2x + \sqrt{1+4x^2}\right).$$\nAt $$t=0$$: $$0 + \frac{1}{2} \ln(0 + 1) = 0.$$\n\n6. **Rewrite the limit:** $$\lim_{x \to \infty} \frac{x \sqrt{1+4x^2} + \frac{1}{2} \ln\left(2x + \sqrt{1+4x^2}\right)}{x^2} = \lim_{x \to \infty} \left( \frac{x \sqrt{1+4x^2}}{x^2} + \frac{\frac{1}{2} \ln\left(2x + \sqrt{1+4x^2}\right)}{x^2} \right).$$\n\n7. **Simplify each term:**\n- For the first term: $$\frac{x \sqrt{1+4x^2}}{x^2} = \frac{\sqrt{1+4x^2}}{x} = \sqrt{\frac{1}{x^2} + 4}.$$ As $$x \to \infty$$, $$\frac{1}{x^2} \to 0$$, so this term approaches $$\sqrt{4} = 2.$$\n- For the second term: $$\frac{\frac{1}{2} \ln\left(2x + \sqrt{1+4x^2}\right)}{x^2}.$$ Since $$\ln(x)$$ grows much slower than any power of $$x$$, this term approaches 0 as $$x \to \infty$$.\n\n8. **Combine the limits:** $$\lim_{x \to \infty} \frac{\int_0^{2x} \sqrt{1+t^2} \, dt}{x^2} = 2 + 0 = 2.$$\n\n**Final answer:** $$\boxed{2}.$$