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

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= Derivatives =
= Derivatives =
<math>\rightarrow</math>


== The derivative of <math>\frac{d}{dx} x^x</math> ==
== The derivative of <math>\frac{d}{dx} x^x</math> ==
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= Limits =
= Limits =


== The limit of <math>\lim_{x \tendsto 0^+} x^x</math> ==
== The limit of <math>\lim_{x \rightarrow 0^+} x^x</math> ==


=== Question ===
=== Question ===


What is <math>\lim_{x \tendsto 0^+} x^x</math>?
What is <math>\lim_{x \rightarrow 0^+} x^x</math>?


=== Solution ===
=== Solution ===

Revision as of 09:46, 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 .
  • 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:

Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto t} f(g(x)) = f(\lim_{x \tendsto t} g(x)). } In our case, is continuous, so Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto 0^+} x^x = e^{\lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}}}. }

  • The question, then, is what is Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}}} .
  • As Failed to parse (unknown function "\tendsto"): {\displaystyle x \tendsto 0^+} , Failed to parse (unknown function "\tendsto"): {\displaystyle \ln x \tendsto -\infty$, $\frac{1}{x} \tendsto +\infty} . In this situation we can apply l'Hôpital's rule:

Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \tendsto 0^+} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} \frac{1}{x}} = \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \frac{\frac{1}{x} \cdot x^2}{-\frac{1}{x^2} \cdot x^2} = -x. }

  • Hence Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto 0^+} x^x = e^0 = 1} .

The limits of Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto +\infty} x \sin \frac{1}{x}} and Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto -\infty} x \sin \frac{1}{x}}

Question

What are Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto +\infty} x \sin \frac{1}{x}} and Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto -\infty} x \sin \frac{1}{x}} ?

Solution

  • Let us rewrite .
  • As Failed to parse (unknown function "\tendsto"): {\displaystyle x \tendsto +\infty$, $\frac{1}{x} \tendsto 0} and Failed to parse (unknown function "\tendsto"): {\displaystyle x \sin \frac{1}{x} \tendsto 0} .
  • We have ``, so we can apply l'Hôpital's rule.
  • Differentiating the numerator in , we obtain .
  • Differentiating the denominator in , we obtain .
  • Thus

Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto +\infty} x \sin \frac{1}{x} = \lim_{x \tendsto +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} = \lim_{x \tendsto +\infty} \frac{\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \tendsto +\infty} \cos \frac{1}{x} = 1. }

  • Similarly we can find that Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto -\infty} x \sin \frac{1}{x} = 1} .