https://one.snvedu.cn/cuoti/Recent content in 错题本 on 专升本备考手册Hugozh-cnSun, 08 Nov 2026 00:00:00 +0000https://one.snvedu.cn/cuoti/2026-11-08-limit-trap/Sun, 08 Nov 2026 00:00:00 +0000https://one.snvedu.cn/cuoti/2026-11-08-limit-trap/<h2 id="原题">原题<a class="anchor" href="#%e5%8e%9f%e9%a2%98" aria-label="Link"></a></h2> <span class="katex"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><munder><mrow><mi>lim</mi><mo>⁡</mo></mrow><mrow><mi>x</mi><mo>→</mo><mn>0</mn></mrow></munder><mfrac><mrow><mi>tan</mi><mo>⁡</mo><mi>x</mi><mo>−</mo><mi>sin</mi><mo>⁡</mo><mi>x</mi></mrow><msup><mi>x</mi><mn>3</mn></msup></mfrac></mrow><annotation encoding="application/x-tex">\lim_{x \to 0}\frac{\tan x-\sin x}{x^3}</annotation></semantics></math></span><h2 id="我的错解">我的错解<a class="anchor" href="#%e6%88%91%e7%9a%84%e9%94%99%e8%a7%a3" aria-label="Link"></a></h2> <p>把 <span class="katex"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>tan</mi><mo>⁡</mo><mi>x</mi></mrow><annotation encoding="application/x-tex">\tan x</annotation></semantics></math></span> 和 <span class="katex"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>sin</mi><mo>⁡</mo><mi>x</mi></mrow><annotation encoding="application/x-tex">\sin x</annotation></semantics></math></span> 都替换成 <span class="katex"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math></span>，分子变 0，得出极限 = 0。❌</p> <h2 id="正确解法">正确解法<a class="anchor" href="#%e6%ad%a3%e7%a1%ae%e8%a7%a3%e6%b3%95" aria-label="Link"></a></h2> <p>先提取公因式把加减变乘除：</p> <span class="katex"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>tan</mi><mo>⁡</mo><mi>x</mi><mo>−</mo><mi>sin</mi><mo>⁡</mo><mi>x</mi><mo>=</mo><mi>tan</mi><mo>⁡</mo><mi>x</mi><mo stretchy="false">(</mo><mn>1</mn><mo>−</mo><mi>cos</mi><mo>⁡</mo><mi>x</mi><mo stretchy="false">)</mo><mo>∼</mo><mi>x</mi><mo>⋅</mo><mfrac><msup><mi>x</mi><mn>2</mn></msup><mn>2</mn></mfrac><mo>=</mo><mfrac><msup><mi>x</mi><mn>3</mn></msup><mn>2</mn></mfrac></mrow><annotation encoding="application/x-tex">\tan x-\sin x=\tan x(1-\cos x)\sim x\cdot\frac{x^2}{2}=\frac{x^3}{2}</annotation></semantics></math></span><p>所以极限 = 1/2。</p>