Compute the limit of <math><mfrac><mrow><mi>sin</mi><mo>(</mo><mn>2</mn><mi>x</mi><mo>)</mo></mrow><mrow><mi>x</mi></mrow></mfrac></math> as <m><math><mi>x</mi><mo> → </mo><mn>0</mn></math></m>
 The limit of <math><mn>2</mn><mi>x</mi></math> is <m><math><mn>0</mn></math></m> when <m><math><mi>x</mi><mo> → </mo><mn>0</mn></math></m>
The limit of <math><mi>sin</mi><mo>(</mo><mn>2</mn><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><mn>2</mn><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><mn>2</mn><mi>x</mi><mo>)</mo></math>
Apply the Taylor development <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> of <m>sin</m> with <m>X</m>=<math><mn>2</mn><mi>x</mi></math> around <m>A</m>=<math><mn>0</mn></math> (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><mn>2</mn><mi>x</mi><mo> + </mo><mn>0</mn><mo>)</mo></math>
Let's restart using the Taylor expansion: <math><mn>2</mn><mi>x</mi></math>
 <math><mfrac><mrow><mn>2</mn><mi>x</mi></mrow><mrow><mi>x</mi></mrow></mfrac></math>
 <math><mn>2</mn></math>
So the limit of <math><mfrac><mrow><mi>sin</mi><mo>(</mo><mn>2</mn><mi>x</mi><mo>)</mo></mrow><mrow><mi>x</mi></mrow></mfrac></math> is <m><math><mn>2</mn></math></m> when <m><math><mi>x</mi><mo> → </mo><mn>0</mn></math></m>

▷<b>Answer: &nbsp; <math><mn>2</mn></math></b><span style='font-size:smaller'> &nbsp; &nbsp; (computation required 11 steps and 2 ms)</span>