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

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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 ===


* Note that <math>x = e^{\ln x}$, so $x^x = (e^{\ln x})^x = e^{x \ln x}</math>.
* Note that <math>x = e^{\ln x}</math>, so <math>x^x = (e^{\ln x})^x = e^{x \ln x}</math>.
* We can further rewrite this as <math>x^x = e^{x \ln x} = e^{\frac{\ln x}{\frac{1}{x}}}</math>.
* We can further rewrite this as <math>x^x = e^{x \ln x} = e^{\frac{\ln x}{\frac{1}{x}}}</math>.
* As long as <math>f</math> is continuous and the limit of <math>g</math> exists at the point in question, the limit will commute with composition:
* As long as <math>f</math> is continuous and the limit of <math>g</math> exists at the point in question, the limit will commute with composition:
<math>
    <math>
\lim_{x \tendsto t} f(g(x)) = f(\lim_{x \tendsto t} g(x)).
    \lim_{x \rightarrow t} f(g(x)) = f(\lim_{x \rightarrow t} g(x)).
</math>
    </math>
In our case, <math>e(\cdot)</math> is continuous, so
In our case, <math>e(\cdot)</math> is continuous, so
<math>
    <math>
\lim_{x \tendsto 0^+} x^x = e^{\lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}}}.
    \lim_{x \rightarrow 0^+} x^x = e^{\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}}.
</math>
    </math>
* The question, then, is what is <math>\lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}}</math>.
* The question, then, is what is <math>\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}</math>.
* As <math>x \tendsto 0^+</math>, <math>\ln x \tendsto -\infty$, $\frac{1}{x} \tendsto +\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 \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.
    \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 \tendsto 0^+} x^x = e^0 = 1</math>.
* Hence <math>\lim_{x \rightarrow 0^+} x^x = e^0 = 1</math>.


== The limits of <math>\lim_{x \tendsto +\infty} x \sin \frac{1}{x}</math> and <math>\lim_{x \tendsto -\infty} x \sin \frac{1}{x}</math> ==
== The limits of <math>\lim_{x \rightarrow +\infty} x \sin \frac{1}{x}</math> and <math>\lim_{x \rightarrow -\infty} x \sin \frac{1}{x}</math> ==


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


What are <math>\lim_{x \tendsto +\infty} x \sin \frac{1}{x}</math> and <math>\lim_{x \tendsto -\infty} x \sin \frac{1}{x}</math>?
What are <math>\lim_{x \rightarrow +\infty} x \sin \frac{1}{x}</math> and <math>\lim_{x \rightarrow -\infty} x \sin \frac{1}{x}</math>?


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


* Let us rewrite <math>x \sin \frac{1}{x}$ as $\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>.
* Let us rewrite <math>x \sin \frac{1}{x}</math> as <math>\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>.
* As <math>x \tendsto +\infty$, $\frac{1}{x} \tendsto 0</math> and <math>x \sin \frac{1}{x} \tendsto 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>.
* Thus
* Thus
<math>
    <math>
\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.
    \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 \tendsto -\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

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 .