سوال ۳۳ کنکور ارشد علوم کامپیوتر ۱۳۹۸

This is a $\tfrac{0}{0}$ form, since $(1+x)^{1/x}\to e$ as $x\to 0$. The clean route is to expand $(1+x)^{1/x}$ to first order in $x$.

Step 1 — Write the power as an exponential.

$(1+x)^{1/x} = \exp\!\left(\frac{\ln(1+x)}{x}\right).$

Step 2 — Expand the exponent. Using $\ln(1+x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \cdots$,

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

Step 3 — Exponentiate. Factor out $e^{1}$ and expand the small part with $e^{u}=1+u+\cdots$:

$(1+x)^{1/x} = e \cdot \exp\!\left(-\frac{x}{2} + O(x^{2})\right) = e\left(1 - \frac{x}{2} + O(x^{2})\right).$

Step 4 — Substitute into the limit.

$e - (1+x)^{1/x} = e - e\left(1 - \frac{x}{2} + O(x^{2})\right) = \frac{e}{2}\,x + O(x^{2}).$

Dividing by $x$:

$\frac{e-(1+x)^{1/x}}{x} = \frac{e}{2} + O(x) \;\xrightarrow[x\to 0]{}\; \frac{e}{2}.$

So the limit equals $\dfrac{e}{2}$, which is option 4.

Exam trick: For limits involving $(1+x)^{1/x}$, immediately write it as $e\,\exp\!\big(-\tfrac{x}{2}+\cdots\big)$ — the key fact is $\frac{\ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^{2}}{3}-\cdots$, so the function approaches $e$ from below at rate $\tfrac{e}{2}$ per unit $x$. The leading correction $-\tfrac{ex}{2}$ is exactly what this problem asks for.