سوال ۲۸ کنکور ارشد علوم کامپیوتر ۱۴۰۲

We must evaluate $L=\lim_{x\to 0^{+}} \frac{\int_{0}^{x} (\sin t)^{\tan t}\, dt}{x e^{x}}.$

Step 1 — Confirm the indeterminate form. The denominator $x e^{x}\to 0$. For the numerator, note the integrand near $t\to 0^{+}$: $(\sin t)^{\tan t}=e^{\tan t\,\ln(\sin t)}.$ Since $\tan t\sim t$ and $\ln(\sin t)\sim \ln t$, the exponent $\tan t\,\ln(\sin t)\sim t\ln t\to 0$, so the integrand $\to 1$. Hence the integrand is bounded near $0$ and $\int_{0}^{x}(\sin t)^{\tan t}\,dt\to 0$. The limit is of the form $\tfrac{0}{0}$.

Step 2 — Apply L'Hôpital with the Fundamental Theorem of Calculus. By the FTC, the derivative of the numerator is the integrand evaluated at $x$: $\frac{d}{dx}\int_{0}^{x} (\sin t)^{\tan t}\, dt=(\sin x)^{\tan x}.$ The denominator differentiates via the product rule: $\frac{d}{dx}\bigl(x e^{x}\bigr)=e^{x}+x e^{x}=e^{x}(1+x).$ Therefore $L=\lim_{x\to 0^{+}} \frac{(\sin x)^{\tan x}}{e^{x}(1+x)}.$

Step 3 — Take the limit. As shown, $(\sin x)^{\tan x}\to 1$, while $e^{x}(1+x)\to e^{0}\cdot 1=1$. Hence $L=\frac{1}{1}=1.$

The answer is 1 (option 2).

Exam trick: Whenever a limit has $\int_{0}^{x} f(t)\,dt$ on top and a function vanishing at $0$ on the bottom, L'Hôpital + FTC collapses the integral to $f(x)$ in one step. Just first verify $f$ is bounded near $0$ so the numerator truly $\to 0$ — here the scary $(\sin x)^{\tan x}$ is the harmless limit $0^{0}\!\to\! 1$, not $0$ or $\infty$.