سوال ۳۱ کنکور ارشد مهندسی کامپیوتر ۱۴۰۰

The base $\cos x + ax^{2}\to 1$ and the exponent $\dfrac{1}{x-\sin x}\to\infty$ as $x\to 0$, so this is a $1^{\infty}$ indeterminate form. Take logarithms:

$\ln A = \lim_{x\to 0}\frac{\ln\!\left(\cos x + ax^{2}\right)}{x-\sin x}.$

Expand the pieces with Taylor series. The denominator is

$x-\sin x = x-\left(x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots\right)=\frac{x^{3}}{6}-\frac{x^{5}}{120}+\cdots,$

so it behaves like $\dfrac{x^{3}}{6}$. For the numerator,

$\cos x + ax^{2}=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\cdots+ax^{2}=1+\left(a-\tfrac12\right)x^{2}+\frac{x^{4}}{24}-\cdots,$

and using $\ln(1+u)\approx u$,

$\ln\!\left(\cos x + ax^{2}\right)\approx\left(a-\tfrac12\right)x^{2}+\frac{x^{4}}{24}-\cdots.$

Force the limit to be bounded. If $a\ne\tfrac12$, the ratio is $\dfrac{(a-\frac12)x^{2}}{x^{3}/6}=\dfrac{6(a-\frac12)}{x}\to\pm\infty$, so $A$ would be $0$ or $\infty$ — not a bounded nonzero number. Hence we must have

$a-\frac12=0\quad\Longrightarrow\quad a=\frac12.$

Evaluate $A$ with $a=\tfrac12$. The $x^{2}$ terms cancel, leaving

$\cos x+\tfrac12 x^{2}=1+\frac{x^{4}}{24}-\cdots,\qquad \ln\!\left(\cos x+\tfrac12 x^{2}\right)\approx\frac{x^{4}}{24}.$

Therefore

$\ln A=\lim_{x\to 0}\frac{x^{4}/24}{x^{3}/6}=\lim_{x\to 0}\frac{6x^{4}}{24x^{3}}=\lim_{x\to 0}\frac{x}{4}=0,$

so $A=e^{0}=1$.

Combine.

$A+a=1+\frac12=\frac{3}{2}.$

Exam trick: In a $1^{\infty}$ limit, the condition “the limit is a bounded (finite, nonzero) number” is a constraint that fixes the unknown parameter — it forces the lowest-order term of $\ln(\text{base})$ to match the order of the denominator. Here matching $x^{3}$ against the $x^{2}$ numerator term kills $(a-\tfrac12)$, then the next term gives $A$.