limite x->0 tan(x)/x
Let's evaluate the limit of <math><mfrac><mrow><mi>tan</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>tan</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>tan</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>tan</mi><mo>(</mo><mi>x</mi><mo>)</mo></math> Apply the Taylor development <math><mi>tan</mi><mo>(</mo><mi>A</mi><mo>)</mo><mo> + </mo><mo>(</mo><mn>1</mn><mo> + </mo><msup><mrow><mi>tan</mi><mo>(</mo><mi>A</mi><mo>)</mo></mrow><mrow><mn>2</mn></mrow></msup><mo>)</mo><mo>·</mo><mo>(</mo><mi>X</mi><mo> -</mo><mi>A</mi><mo>)</mo></math> of <m>tan</m> with <m>X</m>=<math><mi>x</mi></math> around <m>A</m>=<math><mn>0</mn></math> (order <m>N</m>=1) <math><mi>tan</mi><mo>(</mo><mn>0</mn><mo>)</mo><mo> + </mo><mo>(</mo><mn>1</mn><mo> + </mo><msup><mrow><mi>tan</mi><mo>(</mo><mn>0</mn><mo>)</mo></mrow><mrow><mn>2</mn></mrow></msup><mo>)</mo><mo>·</mo><mo>(</mo><mi>x</mi><mo> + </mo><mn>0</mn><mo>)</mo></math> <math><mo>(</mo><mn>1</mn><mo> + </mo><msup><mrow><mn>0</mn></mrow><mrow><mn>2</mn></mrow></msup><mo>)</mo><mo>·</mo><mi>x</mi></math> We redo the calculation 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>tan</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 required 11 steps and 1 ms)</span>
