سوال ۲۷ کنکور ارشد علوم کامپیوتر ۱۴۰۵

Substitute $u=\ln x$. As $x\to+\infty$ we have $u\to+\infty$, so the limit becomes

$L=\lim_{u\to+\infty} u\left(\tan^{-1}u-\frac{\pi}{2}\right).$

This is an $\infty\cdot 0$ indeterminate form, since $\tan^{-1}u\to\frac{\pi}{2}$.

Use the complementary-angle identity. For $u>0$,

$\tan^{-1}u+\tan^{-1}\!\frac{1}{u}=\frac{\pi}{2}\quad\Longrightarrow\quad \tan^{-1}u-\frac{\pi}{2}=-\tan^{-1}\!\frac{1}{u}.$

Therefore

$L=\lim_{u\to+\infty} u\left(-\tan^{-1}\!\frac{1}{u}\right)=-\lim_{u\to+\infty}\frac{\tan^{-1}(1/u)}{1/u}.$

Put $t=\dfrac{1}{u}\to 0^{+}$. Using the standard limit $\dfrac{\tan^{-1}t}{t}\to 1$,

$L=-\lim_{t\to 0^{+}}\frac{\tan^{-1}t}{t}=-1.$

Cross-check via L'Hôpital. Writing $L=\lim_{u\to\infty}\dfrac{\tan^{-1}u-\pi/2}{1/u}$ (form $\tfrac{0}{0}$) and differentiating,

$L=\lim_{u\to\infty}\frac{\dfrac{1}{1+u^{2}}}{-\dfrac{1}{u^{2}}}=\lim_{u\to\infty}\frac{-u^{2}}{1+u^{2}}=-1.$

So the value is $-1$, which is option 3.

Exam trick: For $u>0$, $\tan^{-1}u-\frac{\pi}{2}=-\tan^{-1}(1/u)\approx-\frac{1}{u}$ as $u\to\infty$; multiplying by $u$ instantly gives $-1$ without any differentiation.