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)
Show Hint
Bound \((\sin e^x)^n\) between -1 and 1, divide numerator and denominator by \(n^2\), then take the limit before integrating.
Step 1: Bound the numerator between two simple expressions.
Since $-1 \le \sin e^x \le 1$ for every real $x$, we get $-1 \le (\sin e^x)^n \le 1$ for every positive integer $n$, regardless of what $e^x$ actually equals on $[0,1]$. Adding $n^2$ to all three parts gives
\[
n^2 - 1 \le n^2 + (\sin e^x)^n \le n^2 + 1.
\]
Step 2: Bound the whole integrand using this.
Dividing through by the positive quantity $7n^2 + x^8$,
\[
\frac{n^2-1}{7n^2+x^8} \le \frac{n^2+(\sin e^x)^n}{7n^2+x^8} \le \frac{n^2+1}{7n^2+x^8}, \quad x \in [0,1].
\]
Step 3: Integrate all three parts over $[0,1]$.
Since $0 \le x^8 \le 1$ on $[0,1]$, we can further simplify the outer bounds by replacing $x^8$ with its extreme values $0$ and $1$:
\[
\frac{n^2-1}{7n^2+1} \le \frac{n^2+(\sin e^x)^n}{7n^2+x^8} \le \frac{n^2+1}{7n^2}, \quad x \in [0,1].
\]
Integrating each part over $x$ from $0$ to $1$ (the outer bounds are constants, so their integrals equal themselves):
\[
\frac{n^2-1}{7n^2+1} \le \int_0^1 \frac{n^2+(\sin e^x)^n}{7n^2+x^8}\,dx \le \frac{n^2+1}{7n^2}.
\]
Step 4: Take $n \to \infty$ on both outer bounds.
\[
\lim_{n\to\infty} \frac{n^2-1}{7n^2+1} = \frac{1}{7}, \qquad \lim_{n\to\infty} \frac{n^2+1}{7n^2} = \frac{1}{7}.
\]
Both outer sequences converge to $\frac{1}{7}$, so by the Sandwich (Squeeze) Theorem, the middle sequence, which is exactly the integral defining $\alpha$, also converges to $\frac{1}{7}$. This gives the same value as before, but reaches it purely through inequalities and the Sandwich Theorem, without invoking the Dominated Convergence Theorem at all.
Step 5: State $\alpha$ and compute $14\alpha$.
\[
\alpha = \frac{1}{7} \implies 14\alpha = 14 \times \frac{1}{7} = 2.
\]
\[ \boxed{14\alpha = 2} \]
