2024/G1/2 - Revision history

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:46:30Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:46, 30 March 2023
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	<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}(\frac{3x^2-x-2}{5x^2+4x+1})\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}=\frac{3-(0)-(0)}{5+(0)+(0)}=\frac{3}{5}</math><br>		<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}(\frac{3x^2-x-2}{5x^2+4x+1})\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}=\frac{3-(0)-(0)}{5+(0)+(0)}=\frac{3}{5}</math><br>
	V.A or a vertical asymptote is when the function becomes infinity such as		V.A or a vertical asymptote is when the function becomes infinity such as
	<math>\lim_{x \to \x}f(x)=\infty</math><br>		<math>\lim_{x \to c}f(x)=\infty</math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:46:04Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:46, 30 March 2023
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	<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>		<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>
	<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}(\frac{3x^2-x-2}{5x^2+4x+1})\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}=\frac{3-(0)-(0)}{5+(0)+(0)}=\frac{3}{5}</math><br>		<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}(\frac{3x^2-x-2}{5x^2+4x+1})\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}=\frac{3-(0)-(0)}{5+(0)+(0)}=\frac{3}{5}</math><br>
			V.A or a vertical asymptote is when the function becomes infinity such as
			<math>\lim_{x \to \x}f(x)=\infty</math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:43:06Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:43, 30 March 2023
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	<br>		<br>
	<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>		<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>
	<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}=\frac{3-(0)-(0)}{5+(0)+(0)}=\frac{3}{5}</math><br>		<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}(\frac{3x^2-x-2}{5x^2+4x+1})\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}=\frac{3-(0)-(0)}{5+(0)+(0)}=\frac{3}{5}</math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:42:01Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:42, 30 March 2023
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	<br>		<br>
	<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>		<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>
	<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}</math><br>		<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}=\frac{3-(0)-(0)}{5+(0)+(0)}=\frac{3}{5}</math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:40:52Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:40, 30 March 2023
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	<br>		<br>
	<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>		<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>
	<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}</math><br>		<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}</math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:39:01Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:39, 30 March 2023
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	This ultimately means that the equation will APPROACH TO ONE it will not ever be on one.		This ultimately means that the equation will APPROACH TO ONE it will not ever be on one.
	<br>		<br>
	<math>\lim_{x \to \infty}\frac{1}{x}=0		<math>\lim_{x \to \infty}\frac{1}{x}=0 \lim_{x \to -\infty}\frac{1}{x}=0 </math><br>
	<br>\lim_{x \to -\infty}\frac{1}{x}=0 </math><br>		<math>\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}=\lim_{x \to \infty}\frac{3x^2-x-2}{5x^2+4x+1}\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}</math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:36:46Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:36, 30 March 2023
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	<br>		<br>
	<math> f(x)=\lim_{x \to \infty}\frac{x^2-1}{x^2+1} =\lim_{x \to \infty}\frac{x^2-1}{x^2+1}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}}=\frac{1-0}{1+0}=1 </math><br>		<math> f(x)=\lim_{x \to \infty}\frac{x^2-1}{x^2+1} =\lim_{x \to \infty}\frac{x^2-1}{x^2+1}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}}=\frac{1-0}{1+0}=1 </math><br>
			This ultimately means that the equation will APPROACH TO ONE it will not ever be on one.
			<br>
			<math>\lim_{x \to \infty}\frac{1}{x}=0
			<br>\lim_{x \to -\infty}\frac{1}{x}=0 </math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:33:15Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:33, 30 March 2023
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	Horizontal Asymptote or H.A		Horizontal Asymptote or H.A
	<br>		<br>
	<math> f(x)=\lim_{x \to \infty}\frac{x^2-1}{x^2+1} =\lim_{x \to \infty}\frac{x^2-1}{x^2+1}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}} </math><br>		<math> f(x)=\lim_{x \to \infty}\frac{x^2-1}{x^2+1} =\lim_{x \to \infty}\frac{x^2-1}{x^2+1}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}}=\frac{1-0}{1+0}=1 </math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:32:29Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:32, 30 March 2023
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	Horizontal Asymptote or H.A		Horizontal Asymptote or H.A
	<br>		<br>
	<math> f(x)=\lim_{x \to \infty}\frac{x^2-1}{x^2+1} =\lim_{x \to \infty}\frac{x^2-1}{x^2+1}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} </math><br>		<math> f(x)=\lim_{x \to \infty}\frac{x^2-1}{x^2+1} =\lim_{x \to \infty}\frac{x^2-1}{x^2+1}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\lim_{x \to \infty}\frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}} </math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==

Jeremyr102216@students.laalliance.org: /* 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES */

2023-03-30T21:31:16Z

2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

← Older revision		Revision as of 21:31, 30 March 2023
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	Horizontal Asymptote or H.A		Horizontal Asymptote or H.A
	<br>		<br>
	<math> \lim_{x \to \infty} </math><br>		<math> f(x)=\lim_{x \to \infty}\frac{x^2-1}{x^2+1} =\lim_{x \to \infty}\frac{x^2-1}{x^2+1}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} </math><br>

	==2.7 DERIVATIVES AND RATES OF CHANGE ==		==2.7 DERIVATIVES AND RATES OF CHANGE ==