Compute the limit of <math><mfrac><mrow><mi>ln</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mi>x</mi></mrow></mfrac></math> as <m><math><mi>x</mi><mo> → </mo><mi>+∞</mi></math></m>
The limit of <math><mi>ln</mi><mo>(</mo><mi>x</mi><mo>)</mo></math> is <m><math><mi>+∞</mi></math></m> when <m><math><mi>x</mi><mo> → </mo><mi>+∞</mi></math></m>
For <math><mfrac><mrow><mi>ln</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mi>x</mi></mrow></mfrac></math>, the limit takes the <u>indeterminate form</u> <math><mfrac><mrow><mi>+∞</mi></mrow><mrow><mi>+∞</mi></mrow></mfrac></math> when <math><mi>x</mi></math> → <math><mi>+∞</mi></math>
Let's apply L'Hôpital's rule (<m>limit of a/b = limit of a'/b'</m>) by differentiating numerator and denominator:
Let's perform the differentiation: <math><mo>(</mo><mi>ln</mi><mo>(</mo><mi>x</mi><mo>)</mo><mo>)'</mo></math>
 <math><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi></mrow></mfrac></math>
Let's differentiate: <math><mo>(</mo><mi>x</mi><mo>)'</mo></math>
 <math><mn>1</mn></math>
With the derivatives of the numerator and the denominator, we obtain a new fraction whose limit can be determined:
The limit of <math><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi></mrow></mfrac></math> is <m><math><mn>0</mn></math>⁺</m> when <m><math><mi>x</mi><mo> → </mo><mi>+∞</mi></math></m>
So the limit of <math><mfrac><mrow><mi>ln</mi><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mi>x</mi></mrow></mfrac></math> is <m><math><mn>0</mn></math>⁺</m> when <m><math><mi>x</mi><mo> → </mo><mi>+∞</mi></math></m>

▷<b>Result: &nbsp; <math><mn>0</mn></math></b><span style='font-size:smaller'> &nbsp; &nbsp; (processing: 11 steps, 2 ms)</span>