函数极限

T1 (8.8)

求 \(\lim\limits_{x \rightarrow +\infty} \left[\dfrac{x ^ {1 + x}}{(1 + x) ^ x} - \dfrac{x}{e}\right]\).

solution(8.8)

\[\begin{aligned} I & = \lim\limits_{x \rightarrow +\infty} \dfrac{ex^{1 + x} - x(1 + x) ^ x}{e(1 + x) ^ x} \\ &= \lim\limits_{x \rightarrow +\infty} \dfrac{x(e - \left(1 + \frac{1}{x}) ^ x\right)}{e(1 + \frac{1}{x}) ^ x} \\ &= \lim\limits_{x \rightarrow +\infty} \dfrac{ex\left(1 - e ^ {x\ln (1 + \frac{1}{x}) - 1}\right)}{e(1 + \frac{1}{x}) ^ x}\\ \left(t = \frac{1}{x}\right)&= \dfrac{\lim\limits_{t \rightarrow 0}\dfrac{t - \ln(1 + t)}{t^2}}{ \lim\limits_{t \rightarrow 0} (1 + t) ^ {\frac{1}{t}}}\\ &= \frac{1}{2e} \end{aligned} \]

T2(8.10)

求 \(\lim\limits_{x\rightarrow 0} \dfrac{\frac{x^2}{2}+1-\sqrt{1+x^2}}{(\cos x - e ^ {x^2})\sin x^2}\).

solution(8.10)

\[\begin{aligned} I &= \lim\limits_{x\rightarrow 0} \dfrac{\frac{x ^ 2}{2} + 1 - 1 - \frac{x ^ 2}{2} + \frac{1}{8}x^4 + o(x ^ 4)}{(1 - \frac{x^2}{2} - 1 - x ^ 2 + o(x ^ 2))x^2} = -\frac{1}{12} \end{aligned} \]

T3(8.10)

求 \(\lim\limits_{x\rightarrow 0} \dfrac{\int_{0} ^ x (x - t)f(x - t) dt}{\int_0^{x^2}f(\frac{t}{x})dt}\)，\(f(x)\) 连续且 \(f(0) \neq 0\).

solution(8.10)

• \(\int_{0} ^ x (x - t)f(x - t) dt = \int_0^x tf(t) dt\)
• \(\int_0^{x^2}f(\frac{t}{x})dt = \int_0^xf(u)d(xt)=x\int_0^x f(t)dt\)

\[\begin{aligned} I &= \lim\limits_{x\rightarrow 0}\dfrac{\int_0^x tf(t)dt}{x\int_0^x f(t)dt}\\ &= \lim\limits_{x\rightarrow 0} \dfrac{xf(x)}{\int_0^xf(t)dt + xf(x)} \\ &= \dfrac{1}{1 + \lim\limits_{x\rightarrow 0} \frac{\int_0^x f(t)dt}{xf(x)}} \end{aligned} \]

\[\begin{aligned} J &= \lim\limits_{x\rightarrow 0} \frac{\int_0^x f(t)dt}{xf(x)} \\ &= \lim\limits_{x\rightarrow 0} \dfrac{\int_0^x f(t)dt - 0}{x - 0} . \lim\limits_{x\rightarrow 0} \frac{1}{f(x)} \\ &= f(0) \times \frac{1}{f(0)} = 1 \end{aligned} \]

\[\Rightarrow I = \frac{1}{2} \]

solution(8.13)

\[I = \lim\limits_{x\rightarrow 0}\dfrac{\int_0^x tf(t)dt}{x\int_0^x f(t)dt} = \lim\limits_{x\rightarrow 0} \dfrac{f(\xi_1) \int_0^x tdt}{xf(\xi_2)(x-0)} = \frac{1}{2} \]

T4(8.10)

\(f(x), g(x)\) 连续且 \(f(x)\neq g(x) (x \neq 0)\)，而 \(\lim\limits_{x\rightarrow 0} f(x) = \lim\limits_{x\rightarrow 0} g(x) = 2\).

求 \(\dfrac{(f(x)) ^ {g(x)} - (g(x)) ^ {g(x)}}{f(x) - g(x)}\).

solution(8.10)

\[\begin{aligned} I &= \lim\limits_{x\rightarrow 0} \dfrac{e ^ {g(x)\ln f(x)} - e ^ {g(x)\ln g(x)}}{f(x) - g(x)}\\ \left(\xi_1 \in (g(x)\ln f(x), g(x)\ln g(x))\right)&= \lim\limits_{x\rightarrow 0} \dfrac{e ^ {\xi_1} g(x)(\ln f(x) - \ln g(x))}{ f(x) - g(x)} \\ \left(\xi_2\in(f(x), g(x))\right)&= \lim\limits_{x\rightarrow 0} \dfrac{e ^ {\xi_1} g(x) \frac{1}{\xi_2}(f(x) - g(x))}{f(x) - g(x)} \\ &= e ^ {2\ln 2} \times 2 \times \frac{1}{2} = 4 \end{aligned} \]

solution(8.13)

\[\begin{aligned} I &= \lim\limits_{x\rightarrow 0} \dfrac{e ^ {g(x)\ln f(x)} - e ^ {g(x)\ln g(x)}}{f(x) - g(x)}\\ &= \lim\limits_{x\rightarrow 0} \dfrac{e ^ {g(x)\ln g(x)}(e ^ {\ln\frac{f(x)}{g(x)}} - 1)}{f(x) - g(x)} \\ &= \lim\limits_{x\rightarrow 0} \dfrac{e ^ {g(x)\ln g(x)}\ln\frac{f(x)}{g(x)}}{f(x) - g(x)} \\ &= \lim\limits_{x\rightarrow 0} \dfrac{e ^ {g(x)\ln g(x)}(\frac{f(x)}{g(x)} - 1)}{f(x) - g(x)} = 4 \end{aligned} \]

T5(8.10)

\(f(x)\) 二阶导函数连续，\(f^{''}(x) > 0, f(0) = f^{'}(0) = 0\)，令 \(u = x ^ 2 + \left(\frac{f(x)}{f^{'}(x)}\right) ^ 2\).

求 \(\lim\limits_{x\rightarrow 0} \dfrac{\sin(f(u))}{f(x)x^2}\).

solution(8.10)

先试算 \(u\) 的极限，大概和 \(x ^ 2\) 同阶，求 \(I_0 = \lim\limits_{x\rightarrow 0} \frac{u}{x ^ 2}\).

\[\begin{aligned} I_0 & = 1 + \lim\limits_{x\rightarrow 0} \left(\frac{f(x)}{f ^ {'}(x) x}\right) ^ 2\\ &= 1 + \lim\limits_{x\rightarrow 0}\left(\dfrac{f(0) + f^{'}(0) x + \frac{f ^ {''}(\xi_1)}{2}x^2}{x\left(f ^ {'}(0) + \frac{f ^ {''}(\xi_2)}{2} x\right)}\right) ^ 2 \\ &= \frac{5}{4} \end{aligned} \]

\(\Rightarrow x \rightarrow0 \Rightarrow u\rightarrow 0 \Rightarrow f(u) \rightarrow 0\).

\[\begin{aligned} I &= \lim\limits_{x\rightarrow 0}\dfrac{f(u)}{f(x)x^2} = \lim\limits_{x\rightarrow 0}\dfrac{f(0) + f^{'}(0) u + \frac{f ^ {''}(\xi_3)}{2}u^2}{x ^2\left(f(0) + f^{'}(0) x + \frac{f ^ {''}(\xi_4)}{2}x^2\right)} = I_0^2 = \frac{25}{16} \end{aligned} \]