高等数学（工本）

<div> <h2>极限的四则运算法则</h2> <p>假设当x趋近于a时，函数f(x)和g(x)的极限存在，则有以下四则运算法则：</p> <ul> <li><strong>加法法则：</strong> \(\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)\)</li> <li><strong>减法法则：</strong> \(\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)\)</li> <li><strong>乘法法则：</strong> \(\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)\)</li> <li><strong>除法法则：</strong> \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}\)，且\(\lim_{x \to a} g(x) \neq 0\)</li> </ul> <h3>例题说明</h3> <p>设\(f(x) = x^2 + 1\)，\(g(x) = 2x - 3\)，求\(\lim_{x \to 2} [f(x) + g(x)]\)。</p> <ol> <li>\(\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 1) = 2^2 + 1 = 5\)</li> <li>\(\lim_{x \to 2} g(x) = \lim_{x \to 2} (2x - 3) = 2 \cdot 2 - 3 = 1\)</li> <li>\(\lim_{x \to 2} [f(x) + g(x)] = \lim_{x \to 2} f(x) + \lim_{x \to 2} g(x) = 5 + 1 = 6\)</li> </ol> <p>因此，\(\lim_{x \to 2} [f(x) + g(x)] = 6\)。</p> </div>

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