Let's solve this problem by calculating the limits step-by-step to find the values of $a$ and $b$ for which the conditions are satisfied.

We are given:

1. $\lim_{x \to 0} \frac{\sqrt{a x^2 + b x + 1} - 1}{x} = 3$

2. $\lim_{x \to \infty} \frac{\sqrt{a x^2 + b x + 1}}{x} = 2$

Step 1: Solving the First Limit

For the limit as $x \to 0$: $\lim_{x \to 0} \frac{\sqrt{a x^2 + b x + 1} - 1}{x}$ To handle this expression, let's use the approximation for small values of $x$ around $x = 0$. The square root can be expanded as follows: $\sqrt{a x^2 + b x + 1} \approx 1 + \frac{a x^2 + b x}{2}$ Thus, $\sqrt{a x^2 + b x + 1} - 1 \approx \frac{a x^2 + b x}{2}$ Substitute this into the limit: $\lim_{x \to 0} \frac{\frac{a x^2 + b x}{2}}{x} = \lim_{x \to 0} \frac{a x + b}{2} = \frac{b}{2}$ We are given that this limit equals 3, so: $\frac{b}{2} = 3 \Rightarrow b = 6$

Step 2: Solving the Second Limit

For the limit as $x \to \infty$: $\lim_{x \to \infty} \frac{\sqrt{a x^2 + b x + 1}}{x}$ Divide the terms inside the square root by $x^2$: $\lim_{x \to \infty} \frac{\sqrt{a + \frac{b}{x} + \frac{1}{x^2}}}{1} = \sqrt{a}$ We are given that this limit equals 2, so: $\sqrt{a} = 2 \Rightarrow a = 4$

Solution

The values of $a$ and $b$ that satisfy both conditions are: $a = 4 \quad \text{and} \quad b = 6$

Would you like a further breakdown or have any questions?

Here are 5 related questions you might explore:

1. How would the solution change if the limit at infinity was different?
2. Can we use L'Hôpital's rule here if we don't use approximations?
3. What other methods could we use to evaluate limits without square roots?
4. How does changing the coefficient of $x^2$ in the square root affect the limits?
5. What if $x$ tends to a negative infinity in the second limit?

Tip: Always check if you can simplify an expression by rationalizing or using a Taylor expansion before applying more complex methods.