Difference between revisions of "Mathematics/Calculus/Corner cases"

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* Let us rewrite <math>x \sin \frac{1}{x}</math> as <math>\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>.
* Let us rewrite <math>x \sin \frac{1}{x}</math> as <math>\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>.
* As <math>x \rightarrow +\infty</math>, <math>\frac{1}{x} \rightarrow 0</math> and <math>x \sin \frac{1}{x} \rightarrow 0</math>.
* As <math>x \rightarrow +\infty</math>, <math>\frac{1}{x} \rightarrow 0</math> and <math>x \sin \frac{1}{x} \rightarrow 0</math>.
* We have ``<math>\frac{0}{0}</math>'', so we can apply l'H&ocirc;pital's rule.
* We have "<math>\frac{0}{0}</math>", so we can apply l'H&ocirc;pital's rule.
* Differentiating the numerator in <math>\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>, we obtain <math>\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)</math>.
* Differentiating the numerator in <math>\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>, we obtain <math>\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)</math>.
* Differentiating the denominator in <math>\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>, we obtain <math>-\frac{1}{x^2}</math>.
* Differentiating the denominator in <math>\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>, we obtain <math>-\frac{1}{x^2}</math>.

Revision as of 09:51, 22 December 2020

Derivatives

The derivative of

Question

What is ?

Solution 1

  • Let .
  • Take of both sides: .
  • Differentiate both sides: .
  • Apply the chain rule on the left-hand side: .
  • Apply the product rule on the right-hand side: .
  • Putting it together, we have .
  • Hence .

Solution 2

  • Note that , so .
  • Applying the chain rule, .
  • Applying the product rule, .
  • Therefore .

Integrals

The integral

Question

What is ?

Solution

  • We can write as .
  • Consider the series expansion of :
   .
  • We can interchange the integration and summation (we can recognize this as a special case of the Fubini/Tonelli theorems) and write
   

Limits

The limit of

Question

What is ?

Solution

  • Note that , so .
  • We can further rewrite this as .
  • As long as is continuous and the limit of exists at the point in question, the limit will commute with composition:

In our case, is continuous, so

  • The question, then, is what is .
  • As , , . In this situation we can apply l'Hôpital's rule:

  • Hence .

The limits of and

Question

What are and ?

Solution

  • Let us rewrite as .
  • As , and .
  • We have "", so we can apply l'Hôpital's rule.
  • Differentiating the numerator in , we obtain .
  • Differentiating the denominator in , we obtain .
  • Thus

  • Similarly we can find that .