Math 22 Limits - Revision history

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-19T14:51:04Z

One-Sided Limits and Unbounded Function

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-14T15:06:31Z

One-Sided Limits and Unbounded Function

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-14T15:06:17Z

One-Sided Limits and Unbounded Function

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-14T15:04:48Z

One-Sided Limits and Unbounded Function

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-14T15:03:38Z

One-Sided Limits and Unbounded Function

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-14T14:51:03Z

One-Sided Limits and Unbounded Function

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-14T14:43:24Z

One-Sided Limits and Unbounded Function

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-14T14:42:39Z

One-Sided Limits and Unbounded Function

Tphan046: /* Techniques for Evaluating Limits */

2020-07-14T14:40:55Z

Techniques for Evaluating Limits

Tphan046: /* One-Sided Limits and Unbounded Function */

2020-07-14T14:40:17Z

One-Sided Limits and Unbounded Function

← Older revision		Revision as of 14:51, 19 July 2020
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	So, <math>\lim_{x\to 1} \frac{-2}{x-1}</math> does not exists		So, <math>\lim_{x\to 1} \frac{-2}{x-1}</math> does not exists
		+
		+	[[Math_22\| '''Return to Topics Page''']]

	'''This page were made by [[Contributors\|Tri Phan]]'''		'''This page were made by [[Contributors\|Tri Phan]]'''

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 When <math>x\to 1^+</math>, so <math>x>1</math>, hence <math>x-1>0</math>. Therefore, <math>\lim_{x\to 1^+} \frac{-2}{x-1}=\frac {\text{negative}}{\text{positive}}=-\infty</math>
-Notice: <math>\lim_{x\to 1^-} \frac{-2}{x-1}\ne\lim_{x\to 1^+} \frac{-2}{x-1}</math>. So, <math>\lim_{x\to 1} \frac{-2}{x-1}</math> does not exists
+Notice: <math>\lim_{x\to 1^-} \frac{-2}{x-1}\ne\lim_{x\to 1^+} \frac{-2}{x-1}</math>.
 '''This page were made by [[Contributors|Tri Phan]]'''

@@ Line 51: / Line 51: @@
 When <math>x\to 1^+</math>, so <math>x>1</math>, hence <math>x-1>0</math>. Therefore, <math>\lim_{x\to 1^+} \frac{-2}{x-1}=\frac {\text{negative}}{\text{positive}}=-\infty</math>
+Notice: <math>\lim_{x\to 1^-} \frac{-2}{x-1}\ne\lim_{x\to 1^+} \frac{-2}{x-1}</math>. So, <math>\lim_{x\to 1} \frac{-2}{x-1}</math> does not exists
 '''This page were made by [[Contributors|Tri Phan]]'''

← Older revision		Revision as of 15:04, 14 July 2020
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	When <math>x\to 1^-</math>, so <math>x<1</math>, hence <math>x-1<0</math>. Therefore, <math>\lim_{x\to 1^-} \frac{-2}{x-1}=\frac {\text{negative}}{\text{negative}}=\infty</math>		When <math>x\to 1^-</math>, so <math>x<1</math>, hence <math>x-1<0</math>. Therefore, <math>\lim_{x\to 1^-} \frac{-2}{x-1}=\frac {\text{negative}}{\text{negative}}=\infty</math>

		+	When <math>x\to 1^+</math>, so <math>x>1</math>, hence <math>x-1>0</math>. Therefore, <math>\lim_{x\to 1^+} \frac{-2}{x-1}=\frac {\text{negative}}{\text{positive}}=-\infty</math>

← Older revision		Revision as of 15:03, 14 July 2020
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	In some case, the limit of <math>f(x)</math> can be increase/decrease without bound as <math>x</math> approaches <math>c</math>. We can write <math>\lim_{x\to c} f(x)=\pm\infty</math>		In some case, the limit of <math>f(x)</math> can be increase/decrease without bound as <math>x</math> approaches <math>c</math>. We can write <math>\lim_{x\to c} f(x)=\pm\infty</math>

−	~~Consider~~ <math>\lim_{x\to 1} \frac {2}{x-1}</math>. By direct substitution, it is of the form <math>\frac {\text{constant}}{0}</math>, so the answer will be either <math>\infty</math> or <math>-\infty</math>	+	Now, consider <math>\lim_{x\to 1} \frac {-2}{x-1}</math>. By direct substitution, it is of the form <math>\frac {\text{constant}}{0}</math>, so the answer will be either <math>\infty</math> or <math>-\infty</math>. In order to find the limit, we must consider the limit from both side (<math>\lim_{x\to 1^-}</math> and <math>\lim_{x\to 1^+}</math>).
		+
		+	When <math>x\to 1^-</math>, so <math>x<1</math>, hence <math>x-1<0</math>. Therefore, <math>\lim_{x\to 1^-} \frac{-2}{x-1}=\frac {\text{negative}}{\text{negative}}=\infty</math>

@@ Line 42: / Line 42: @@
 One-sided Limit is related to unbounded function.
 Consider <math>\lim_{x\to 1} \frac {2}{x-1}</math>. By direct substitution, it is of the form <math>\frac {\text{constant}}{0}</math>, so the answer will be either <math>\infty</math> or <math>-\infty</math>
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-'''This page is under constuction'''
 '''This page were made by [[Contributors|Tri Phan]]'''

← Older revision		Revision as of 14:43, 14 July 2020
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	One-sided Limit is related to unbounded function.		One-sided Limit is related to unbounded function.

−	Consider <math>\lim_{x\to 1} \frac {2}{x-1}</math>. By direct substitution, it is of the form <math>\frac {text{constant}}{0}</math>	+	Consider <math>\lim_{x\to 1} \frac {2}{x-1}</math>. By direct substitution, it is of the form <math>\frac {\text{constant}}{0}</math>, so the answer will be either <math>\infty</math> or <math>-\infty</math>

← Older revision		Revision as of 14:42, 14 July 2020
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	One-sided Limit is related to unbounded function.		One-sided Limit is related to unbounded function.

−	Consider <math>\lim_{x\to 1} \frac {2}{x-1}</math>	+	Consider <math>\lim_{x\to 1} \frac {2}{x-1}</math>. By direct substitution, it is of the form <math>\frac {text{constant}}{0}</math>

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 '''1. Direct Substitution''': Direct Substitution can be used to find the limit of a Polynomial Function.
-Example: Evaluate <math>lim_{x\to 3} x^2+2x-1=(3)^2+2(3)-1=14</math>
+Example: Evaluate <math>\lim_{x\to 3} x^2+2x-1=(3)^2+2(3)-1=14</math>
 '''2. Dividing Out Technique''': When direct substitution fails and numerator or/and denominator can be factored.
-Example: Evaluate <math>lim_{x\to 2} \frac {x^2-4}{x^2-x-2}= lim_{x\to 2} \frac {(x-2)(x+2)}{(x-2)(x+1)}=lim_{x\to 2} \frac {x+2}{x+1}</math>. Now we can use direct substitution to get the answer.
+Example: Evaluate <math>\lim_{x\to 2} \frac {x^2-4}{x^2-x-2}= \lim_{x\to 2} \frac {(x-2)(x+2)}{(x-2)(x+1)}=\lim_{x\to 2} \frac {x+2}{x+1}</math>. Now we can use direct substitution to get the answer.
 '''3. Rationalizing (Using Conjugate)''': When direct substitution fails and either numerator or denominator has a square root. In this case, we can try to multiply both numerator and denominator by the conjugate.
-Example: Evaluate <math>lim_{x\to 0} \frac {\sqrt{x+4}-2}{x}= lim_{x\to 0} \frac {\sqrt{x+4}-2} {x} \cdot \frac {\sqrt{x+4}+2}{\sqrt{x+4}+2}= lim_{x\to 0} \frac {(x+4)-4}{x(\sqrt{x+4}+2)}=lim_{x\to 0} \frac {x}{x(\sqrt{x+4}+2)}=lim_{x\to 0} \frac {1}{\sqrt{x+4}+2}</math>. Now we can use direct substitution to get the answer
+Example: Evaluate <math>\lim_{x\to 0} \frac {\sqrt{x+4}-2}{x}= \lim_{x\to 0} \frac {\sqrt{x+4}-2} {x} \cdot \frac {\sqrt{x+4}+2}{\sqrt{x+4}+2}= \lim_{x\to 0} \frac {(x+4)-4}{x(\sqrt{x+4}+2)}=\lim_{x\to 0} \frac {x}{x(\sqrt{x+4}+2)}=\lim_{x\to 0} \frac {1}{\sqrt{x+4}+2}</math>. Now we can use direct substitution to get the answer
 ==One-Sided Limits and Unbounded Function==

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    when a function approaches a different value from the left of <math>c</math> than it approaches from the right of <math>c</math>, the limit does not exists. However, this type of behavior can be described more concisely with
    the concept of a one-sided limit. We denote
-   <math>lim_{x\to c^{-}} f(x)=L</math> and <math>lim_{x\to c^{+}} f(x)=K</math>
+   <math>\lim_{x\to c^{-}} f(x)=L</math> and <math>\lim_{x\to c^{+}} f(x)=K</math>
 One-sided Limit is related to unbounded function.