Question

Let \[ \alpha = \lim_{n \to \infty} \int_0^{1} \frac{n^2 + (\sin e^x)^n}{7n^2 + x^8} \, dx. \] The value of \(14\alpha\) is equal to ______. (Answer in integer)

Hint:

Bound \((\sin e^x)^n\) between -1 and 1, divide numerator and denominator by \(n^2\), then take the limit before integrating.

Answer 1

Correct Answer: 2

Step 1: Look at the two parts of the integrand separately.
The integrand is \( \dfrac{n^2 + (\sin e^x)^n}{7n^2 + x^8} \) on \(x \in [0,1]\). As \(n \to \infty\), the term \(n^2\) grows without bound, while \((\sin e^x)^n\) stays bounded because \(|\sin e^x| \leq 1\) for every real \(x\), so \(|(\sin e^x)^n| \leq 1\) for every \(n\).

Step 2: Divide numerator and denominator by \(n^2\).
Write the integrand as \[ \frac{n^2 + (\sin e^x)^n}{7n^2 + x^8} = \frac{1 + \dfrac{(\sin e^x)^n}{n^2}}{7 + \dfrac{x^8}{n^2}}. \] For each fixed \(x \in [0,1]\), the term \(\dfrac{(\sin e^x)^n}{n^2}\) goes to \(0\) as \(n \to \infty\) because the numerator stays between \(-1\) and \(1\) while the denominator \(n^2\) grows without bound. In the same way \(\dfrac{x^8}{n^2} \to 0\) since \(x^8 \leq 1\) on \([0,1]\).

Step 3: Find the pointwise limit of the integrand.
So for every \(x \in [0,1]\), \[ \lim_{n \to \infty} \frac{n^2 + (\sin e^x)^n}{7n^2 + x^8} = \frac{1+0}{7+0} = \frac{1}{7}. \] The whole sequence of functions converges pointwise to the constant \(\dfrac{1}{7}\) on \([0,1]\).

Step 4: Check that we can move the limit inside the integral.
For \(n \geq 1\), the numerator \(n^2 + (\sin e^x)^n\) is at most \(n^2+1\), and the denominator \(7n^2+x^8\) is at least \(7n^2\), so the integrand is bounded by \(\dfrac{n^2+1}{7n^2} \leq \dfrac{2}{7}\), a constant that does not depend on \(x\). Since this bound is an integrable function on the finite interval \([0,1]\), the Dominated Convergence Theorem lets us swap the limit and the integral.

Step 5: Compute \(\alpha\) and then \(14\alpha\).
\[ \alpha = \lim_{n\to\infty}\int_0^1 \frac{n^2+(\sin e^x)^n}{7n^2+x^8}\,dx = \int_0^1 \frac{1}{7}\,dx = \frac{1}{7}\left[x\right]_0^1 = \frac{1}{7}. \] So \[ 14\alpha = 14 \times \frac{1}{7} = 2. \]
Final Answer:
The value of \(14\alpha\) is 2. \[ \boxed{14\alpha = 2} \]