Tìm tiệm cận của hàm số: \(f(x) = \frac{x}{{1 + {e^{\frac{1}{x}}}}}\)

To find the oblique asymptote of the function \(f(x) = \frac{x}{{1 + {e^{\frac{1}{x}}}}}\) , we need to calculate the following limit: \(a = \mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{x}{{1 + {e^{\frac{1}{x}}}}}}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{1 + {e^{\frac{1}{x}}}}}\) When \(x \to \infty \), then \(\frac{1}{x} \to 0\), so \({e^{\frac{1}{x}}} \to {e^0} = 1\). Thus, \(a = \frac{1}{{1 + 1}} = \frac{1}{2}\). Next, we calculate: \(b = \mathop {\lim }\limits_{x \to \infty } \left( {f(x) - ax} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{x}{{1 + {e^{\frac{1}{x}}}}} - \frac{1}{2}x} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{2x - x(1 + {e^{\frac{1}{x}}})}}{{2(1 + {e^{\frac{1}{x}}})}} = \mathop {\lim }\limits_{x \to \infty } \frac{{x - x{e^{\frac{1}{x}}}}}{{2(1 + {e^{\frac{1}{x}}})}} = \mathop {\lim }\limits_{x \to \infty } \frac{{x(1 - {e^{\frac{1}{x}}})}}{{2(1 + {e^{\frac{1}{x}}})}}\) Let \(t = \frac{1}{x}\), when \(x \to \infty \) then \(t \to 0\), we have: \(b = \mathop {\lim }\limits_{t \to 0} \frac{{\frac{1}{t}(1 - {e^t})}}{{2(1 + {e^t})}} = \mathop {\lim }\limits_{t \to 0} \frac{{1 - {e^t}}}{{2t(1 + {e^t})}}\) Applying L'Hopital's rule: \(b = \mathop {\lim }\limits_{t \to 0} \frac{{ - {e^t}}}{{2(1 + {e^t}) + 2t{e^t}}} = \frac{{ - {e^0}}}{{2(1 + {e^0}) + 2 \cdot 0 \cdot {e^0}}} = \frac{{ - 1}}{{2(1 + 1)}} = - \frac{1}{4}\) Thus, the oblique asymptote of the function is \(y = ax + b = \frac{x}{2} - \frac{1}{4}\).