Mathematics/Calculus/Corner cases - Revision history

Admin at 09:53, 22 December 2020

2020-12-22T09:53:31Z

← Older revision		Revision as of 09:53, 22 December 2020
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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'Hôpital'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'Hôpital'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>.

Admin at 09:51, 22 December 2020

2020-12-22T09:51:25Z

← Older revision		Revision as of 09:51, 22 December 2020
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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'Hôpital's rule.		* We have "<math>\frac{0}{0}</math>", so we can apply l'Hôpital'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>.

Admin at 09:50, 22 December 2020

2020-12-22T09:50:40Z

← Older revision		Revision as of 09:50, 22 December 2020
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	=== 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}</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 \~~tendsto~~ +\infty</math>, <math>\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'Hôpital's rule.		* We have ``<math>\frac{0}{0}</math>'', so we can apply l'Hôpital'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>.
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	* 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>.

Admin at 09:49, 22 December 2020

2020-12-22T09:49:29Z

← Older revision		Revision as of 09:49, 22 December 2020
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	=== Solution ===		=== Solution ===

	* Note that <math>x = e^{\ln x}$, so $x^x = (e^{\ln x})^x = e^{x \ln x}</math>.		* Note that <math>x = e^{\ln x}</math>, so <math>x^x = (e^{\ln x})^x = e^{x \ln x}</math>.
	* We can further rewrite this as <math>x^x = e^{x \ln x} = e^{\frac{\ln x}{\frac{1}{x}}}</math>.		* We can further rewrite this as <math>x^x = e^{x \ln x} = e^{\frac{\ln x}{\frac{1}{x}}}</math>.
	* As long as <math>f</math> is continuous and the limit of <math>g</math> exists at the point in question, the limit will commute with composition:		* As long as <math>f</math> is continuous and the limit of <math>g</math> exists at the point in question, the limit will commute with composition:
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	</math>		</math>
	* The question, then, is what is <math>\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}</math>.		* The question, then, is what is <math>\lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}}</math>.
	* As <math>x \rightarrow 0^+</math>, <math>\ln x \rightarrow -\infty$, $\frac{1}{x} \rightarrow +\infty</math>. In this situation we can apply l'Hôpital'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'Hôpital'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.
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	=== 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 \tendsto +\infty</math>, <math>\frac{1}{x} \tendsto 0</math> and <math>x \sin \frac{1}{x} \tendsto 0</math>.
	* We have ``<math>\frac{0}{0}</math>'', so we can apply l'Hôpital's rule.		* We have ``<math>\frac{0}{0}</math>'', so we can apply l'Hôpital'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>.

Admin at 09:48, 22 December 2020

2020-12-22T09:48:29Z

← Older revision		Revision as of 09:48, 22 December 2020
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	\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 \~~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 ===

Admin at 09:48, 22 December 2020

2020-12-22T09:48:01Z

← Older revision		Revision as of 09:48, 22 December 2020
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	\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 \~~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$, $\frac{1}{x} \~~tendsto~~ +\infty</math>. In this situation we can apply l'Hôpital's rule:		* As <math>x \rightarrow 0^+</math>, <math>\ln x \rightarrow -\infty$, $\frac{1}{x} \rightarrow +\infty</math>. In this situation we can apply l'Hôpital'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 \tendsto 0^+} x^x = e^0 = 1</math>.

Admin at 09:46, 22 December 2020

2020-12-22T09:46:59Z

← Older revision		Revision as of 09:46, 22 December 2020
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	* 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 \tendsto 0^+} \frac{\ln x}{\frac{1}{x}}</math>.

Admin at 09:46, 22 December 2020

2020-12-22T09:46:12Z

← Older revision		Revision as of 09:46, 22 December 2020
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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 ===

Admin: Test edit.

2020-12-22T09:45:40Z

Test edit.

← Older revision		Revision as of 09:45, 22 December 2020
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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> ==

Admin: Added some corner cases.

2020-12-22T09:19:51Z

Added some corner cases.

← Older revision		Revision as of 09:19, 22 December 2020
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	= Limits =		= Limits =

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

			=== Question ===

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

			=== Solution ===

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

			=== 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>?

			=== Solution ===

			* Let us rewrite <math>x \sin \frac{1}{x}$ as $\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>.
			* As <math>x \tendsto +\infty$, $\frac{1}{x} \tendsto 0</math> and <math>x \sin \frac{1}{x} \tendsto 0</math>.
			* We have ``<math>\frac{0}{0}</math>'', so we can apply l'Hôpital'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 denominator in <math>\frac{\sin \frac{1}{x}}{\frac{1}{x}}</math>, we obtain <math>-\frac{1}{x^2}</math>.
			* Thus
			<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.
			</math>
			* Similarly we can find that <math>\lim_{x \tendsto -\infty} x \sin \frac{1}{x} = 1</math>.