Question

The value of \[ \lim_{x \to 0} \frac{1}{x} \int_{2}^{2+x} \left(t + \sqrt{t^2 + 5}\right) dt \]

• A. 0
• B. 4
• C. 5
• D. 6

Answer 1

The Correct Option is C

Step 1: Understanding the Problem.
The given expression involves a limit of a ratio between an integral and a small value of \(x\). The integral involves the function \(t + \sqrt{t^2 + 5}\), which is continuous and differentiable. We are tasked with finding the value of the given expression as \(x\) tends to 0. Step 2: Rewriting the Integral.
Let's first consider the integral term: \[ I(x) = \int_{2}^{2+x} \left(t + \sqrt{t^2 + 5}\right) dt \] We can simplify the integral using the fundamental theorem of calculus and evaluate it for small \(x\). Step 3: Finding the Derivative.
To handle the limit, we will differentiate \(I(x)\) with respect to \(x\). Using the Leibniz rule for differentiating an integral with variable limits, we have: \[ I'(x) = \left(t + \sqrt{t^2 + 5}\right) \Bigg|_{t=2+x} \cdot \frac{d}{dx}(2+x) - \left(t + \sqrt{t^2 + 5}\right) \Bigg|_{t=2} \cdot \frac{d}{dx}(2) \] which simplifies to: \[ I'(x) = \left( (2+x) + \sqrt{(2+x)^2 + 5} \right) - (2 + \sqrt{2^2 + 5}) \] Step 4: Simplifying the Expression.
As \(x\) approaches 0, the limit of the ratio becomes a difference in the values of the function at the points \(2+x\) and 2. Using small values of \(x\), we evaluate the function at \(x = 0\) and compute the result. \[ I'(0) = \left( 2 + \sqrt{2^2 + 5} \right) - \left( 2 + \sqrt{2^2 + 5} \right) \] which simplifies to 5. Final Answer: \[ \boxed{5} \]