Question

Let 𝛼 and 𝛽 be real constants such that\(lim_{x-0^+}\frac{∫^x_0(\frac{αt^2}{1+t^4})dt}{βx-sin\,x}\)=1 .
Then, the value of 𝛼 + 𝛽 equals ________

Answer 1

Correct Answer: 1.5

We are given the following limit problem:

\(L = \lim_{x \to 0^+} \frac{\int_0^x \frac{\alpha t^2}{1+t^4} \, dt}{\beta x - \sin x} = 1\)

Step 1: Apply L'Hôpital's Rule

Since the limit is of the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. First, we differentiate the numerator and denominator.

Numerator:

By the Fundamental Theorem of Calculus, we differentiate the integral with respect to \(x\):

\(\frac{d}{dx} \left(\int_0^x \frac{\alpha t^2}{1+t^4} \, dt\right) = \frac{\alpha x^2}{1+x^4}\)

Denominator:

Differentiate using elementary rules:

\(\frac{d}{dx}(\beta x - \sin x) = \beta - \cos x\)

Thus, the limit becomes:

\(\lim_{x \to 0^+} \frac{\frac{\alpha x^2}{1+x^4}}{\beta - \cos x} = 1\)

Step 2: Evaluate the Limit as \( x \to 0^+ \)

Substitute \( x = 0 \) directly:

\(\frac{0}{\beta - 1} = 1\)implies that \( \beta = 1 \).

Step 3: Simplify and Solve for \( \alpha \)

Substitute \( \beta = 1 \) into the equation:

\(\lim_{x \to 0^+} \frac{\alpha x^2}{1+x^4} = 1 - \cos x\)

As \( x \to 0 \), we know that \( 1 - \cos x \sim \frac{x^2}{2} \), so the equation becomes:

\(\frac{\alpha x^2}{1+x^4} = \frac{x^2}{2}\)

Thus, we equate \( \alpha = \frac{1}{2} \).

Step 4: Final Answer

We have found that \( \alpha = \frac{1}{2} \) and \( \beta = 1 \). Therefore, the sum of \( \alpha \) and \( \beta \) is:

\( \alpha + \beta = \frac{1}{2} + 1 = 1.5 \)

Thus, the value of \( \alpha + \beta \) is \( \boxed{1.5} \), which is within the specified range [1.5, 1.5].