سوال ۳۱ کنکور ارشد علوم کامپیوتر ۱۳۹۹

The base tends to $0^{+}$ and the exponent $\tfrac{1}{\ln|x|}\to 0^{-}$, so this is a $0^{0}$ indeterminate form. Take logarithms and let $L=\lim_{x\to 0}\dfrac{\ln(\text{base})}{\ln|x|}$; then the answer is $e^{L}$.

Simplify the base. Pull $e$ out of the logarithm:

$\frac{x+e}{e} - \ln(x+e) = 1+\frac{x}{e} - \ln\!\left(e\Big(1+\frac{x}{e}\Big)\right) = 1+\frac{x}{e} - 1 - \ln\!\left(1+\frac{x}{e}\right) = \frac{x}{e} - \ln\!\left(1+\frac{x}{e}\right).$

Taylor-expand. With $u=\dfrac{x}{e}\to 0$, use $\ln(1+u)=u-\dfrac{u^{2}}{2}+O(u^{3})$:

$u - \ln(1+u) = \frac{u^{2}}{2} + O(u^{3}) = \frac{x^{2}}{2e^{2}} + O(x^{3}).$

So the base behaves like $\dfrac{x^{2}}{2e^{2}}$ near $0$ (positive on both sides, consistent with $\ln|x|$).

Take the log and divide by $\ln|x|$.

$\ln(\text{base}) = \ln\!\left(\frac{x^{2}}{2e^{2}}\right) + o(1) = 2\ln|x| - \ln(2e^{2}) + o(1).$

Therefore

$L = \lim_{x\to 0}\frac{2\ln|x| - \ln(2e^{2})}{\ln|x|} = \lim_{x\to 0}\left(2 - \frac{\ln(2e^{2})}{\ln|x|}\right) = 2,$

since $\ln|x|\to -\infty$ kills the constant term.

Conclusion. The limit equals $e^{L}=e^{2}$, which is option 4.

Exam trick: For a $0^{0}$ (or $1^{\infty}$) limit, take $\ln$ and the only thing that survives in $\dfrac{\ln(\text{base})}{\ln|x|}$ is the leading power of the base. Here base $\sim \tfrac{x^{2}}{2e^{2}}$, so $\ln(\text{base})\sim 2\ln|x|$ and the ratio $\to 2$ — the messy constant $\tfrac{1}{2e^{2}}$ is irrelevant because $\dfrac{\text{const}}{\ln|x|}\to 0$.