limit x->0 sin(x)/x
Let's evaluate the limit of <math><mfrac><mrow><mi>sin</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mi>x</mi></mrow></mfrac></math> when <m><math><mi>x</mi><mo> → </mo><mn>0</mn></math></m> The limit of <math><mi>sin</mi><mo>(</mo><mi>x</mi><mo>)</mo></math> is <m><math><mn>0</mn></math></m> when <m><math><mi>x</mi><mo> → </mo><mn>0</mn></math></m> For <math><mfrac><mrow><mi>sin</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mi>x</mi></mrow></mfrac></math>, we encounter the <u>indeterminate form</u> <m>0/0</m> as <math><mi>x</mi></math> tends to <math><mn>0</mn></math> We look for a Taylor expansion of the expression: <math><mi>sin</mi><mo>(</mo><mi>x</mi><mo>)</mo></math> Using Taylor's formula: <math><mi>sin</mi><mo>(</mo><mi>A</mi><mo>)</mo><mo> + </mo><mi>cos</mi><mo>(</mo><mi>A</mi><mo>)</mo><mo>·</mo><mo>(</mo><mi>X</mi><mo> -</mo><mi>A</mi><mo>)</mo></math> for the function <m>sin</m>, we compute the Taylor expansion for <m>X</m>=<math><mi>x</mi></math> at the point <m>A</m>=<math><mn>0</mn></math> of order <m>N</m>=1 <math><mi>sin</mi><mo>(</mo><mn>0</mn><mo>)</mo><mo> + </mo><mi>cos</mi><mo>(</mo><mn>0</mn><mo>)</mo><mo>·</mo><mo>(</mo><mi>x</mi><mo> + </mo><mn>0</mn><mo>)</mo></math> Let's restart using the Taylor expansion: <math><mi>x</mi></math> <math><mfrac><mrow><mi>x</mi></mrow><mrow><mi>x</mi></mrow></mfrac></math> <math><mn>1</mn></math> So the limit of <math><mfrac><mrow><mi>sin</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mi>x</mi></mrow></mfrac></math> is <m><math><mn>1</mn></math></m> when <m><math><mi>x</mi><mo> → </mo><mn>0</mn></math></m> ▷<b>Result: <math><mn>1</mn></math></b><span style='font-size:smaller'> (computation took 10 steps and 2 ms)</span>
