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

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* We can further rewrite this as <math>x^x = e^{x \ln x} = e^{\frac{\ln x}{\frac{1}{x}}}</math>.
* We can further rewrite this as <math>x^x = e^{x \ln x} = e^{\frac{\ln x}{\frac{1}{x}}}</math>.
* As long as <math>f</math> is continuous and the limit of <math>g</math> exists at the point in question, the limit will commute with composition:
* As long as <math>f</math> is continuous and the limit of <math>g</math> exists at the point in question, the limit will commute with composition:
<math>
    <math>
\lim_{x \rightarrow t} f(g(x)) = f(\lim_{x \rightarrow t} g(x)).
    \lim_{x \rightarrow t} f(g(x)) = f(\lim_{x \rightarrow t} g(x)).
</math>
    </math>
In our case, <math>e(\cdot)</math> is continuous, so
In our case, <math>e(\cdot)</math> is continuous, so
<math>
    <math>
\lim_{x \rightarrow 0^+} x^x = e^{\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}}.
    \lim_{x \rightarrow 0^+} x^x = e^{\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}}.
</math>
    </math>
* The question, then, is what is <math>\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}</math>.
* The question, then, is what is <math>\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}</math>.
* As <math>x \rightarrow 0^+</math>, <math>\ln x \rightarrow -\infty</math>, <math>\frac{1}{x} \rightarrow +\infty</math>. In this situation we can apply l'H&ocirc;pital's rule:
* As <math>x \rightarrow 0^+</math>, <math>\ln x \rightarrow -\infty</math>, <math>\frac{1}{x} \rightarrow +\infty</math>. In this situation we can apply l'H&ocirc;pital's rule:
<math>
    <math>
\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \rightarrow 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.
    \lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \rightarrow 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.
</math>
    </math>
* Hence <math>\lim_{x \rightarrow 0^+} x^x = e^0 = 1</math>.
* Hence <math>\lim_{x \rightarrow 0^+} x^x = e^0 = 1</math>.


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* 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>.
* Thus
* Thus
<math>
    <math>
\lim_{x \rightarrow +\infty} x \sin \frac{1}{x} = \lim_{x \rightarrow +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} = \lim_{x \rightarrow +\infty} \frac{\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \rightarrow +\infty} \cos \frac{1}{x} = 1.
    \lim_{x \rightarrow +\infty} x \sin \frac{1}{x} = \lim_{x \rightarrow +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} = \lim_{x \rightarrow +\infty} \frac{\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \rightarrow +\infty} \cos \frac{1}{x} = 1.
</math>
    </math>
* Similarly we can find that <math>\lim_{x \rightarrow -\infty} x \sin \frac{1}{x} = 1</math>.
* Similarly we can find that <math>\lim_{x \rightarrow -\infty} x \sin \frac{1}{x} = 1</math>.

Latest revision as of 09:53, 22 December 2020

Derivatives

The derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x^x}

Question

What is ?

Solution 1

  • Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = x^x} .
  • Take Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln} of both sides: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln y = x \ln x} .
  • Differentiate both sides: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} \ln y = \frac{d}{dx} x \ln x} .
  • Apply the chain rule on the left-hand side: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} \ln y = \frac{1}{y} \cdot \frac{dy}{dx}} .
  • Apply the product rule on the right-hand side: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x \ln x = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1} .
  • Putting it together, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1} .
  • Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1)} .

Solution 2

  • Note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = e^{\ln x}} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^x = (e^{\ln x})^x = e^{x \ln x} } .
  • Applying the chain rule, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x^x = \frac{d}{dx} e^{x \ln x} = e^{x \ln x} \frac{d}{dx} x \ln x} .
  • Applying the product rule, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x \ln x = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1} .
  • Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx} x^x = e^{x \ln x} (\ln x + 1) = x^x (\ln x + 1)} .

Integrals

The integral Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^x \, dx}

Question

What is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^x \, dx} ?

Solution

  • We can write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^x} as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (e^{\ln x})^x = e^{x \ln x}} .
  • Consider the series expansion of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{x \ln x}} :
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{x \ln x} = 1 + (x \ln x) + \frac{(x \ln x)^2}{2!} + \frac{(x \ln x)^3}{3!} + \ldots + \frac{(x \ln x)^i}{i!} + \ldots = \sum_{i=0}^{\infty} \frac{(x \ln x)^i}{i!}}
.
  • We can interchange the integration and summation (we can recognize this as a special case of the Fubini/Tonelli theorems) and write
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle      \int x^x \, dx = \int \left( \sum_{i=0}^{\infty} \frac{(x \ln x)^i}{i!} \right) \, dx = \sum_{i=0}^{\infty} \left( \int \frac{(x \ln x)^i}{i!} \, dx \right) = \sum_{i=0}^{\infty} \left( \frac{1}{i!} \int x^i (\ln x)^i \, dx \right).     }

Limits

The limit of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow 0^+} x^x}

Question

What is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow 0^+} x^x} ?

Solution

  • Note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = e^{\ln x}} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^x = (e^{\ln x})^x = e^{x \ln x}} .
  • We can further rewrite this as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^x = e^{x \ln x} = e^{\frac{\ln x}{\frac{1}{x}}}} .
  • As long as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous and the limit of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} exists at the point in question, the limit will commute with composition:
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle      \lim_{x \rightarrow t} f(g(x)) = f(\lim_{x \rightarrow t} g(x)).     }

In our case, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e(\cdot)} is continuous, so 

   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle      \lim_{x \rightarrow 0^+} x^x = e^{\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}}.     }
  • The question, then, is what is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}} .
  • As Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \rightarrow 0^+} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln x \rightarrow -\infty} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{x} \rightarrow +\infty} . In this situation we can apply l'Hôpital's rule:
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle      \lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \rightarrow 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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow 0^+} x^x = e^0 = 1} .

The limits of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow +\infty} x \sin \frac{1}{x}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow -\infty} x \sin \frac{1}{x}}

Question

What are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow +\infty} x \sin \frac{1}{x}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow -\infty} x \sin \frac{1}{x}} ?

Solution

  • Let us rewrite Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \sin \frac{1}{x}} as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\sin \frac{1}{x}}{\frac{1}{x}}} .
  • As Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \rightarrow +\infty} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{x} \rightarrow 0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \sin \frac{1}{x} \rightarrow 0} .
  • We have "Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{0}{0}} ", so we can apply l'Hôpital's rule.
  • Differentiating the numerator in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\sin \frac{1}{x}}{\frac{1}{x}}} , we obtain Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)} .
  • Differentiating the denominator in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\sin \frac{1}{x}}{\frac{1}{x}}} , we obtain Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{1}{x^2}} .
  • Thus
   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle      \lim_{x \rightarrow +\infty} x \sin \frac{1}{x} = \lim_{x \rightarrow +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} = \lim_{x \rightarrow +\infty} \frac{\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \rightarrow +\infty} \cos \frac{1}{x} = 1.     }
  • Similarly we can find that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \rightarrow -\infty} x \sin \frac{1}{x} = 1} .
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