Math 22 Limits

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The Limit of a Function

 Definition of the Limit of a Function
 If  becomes arbitrarily close to a single number  as  approaches  from either side, then
 
 which is read as "the limit of  as  approaches  is 

Note: Many times the limit of as approaches is simply , so limit can be evaluate by direct substitution as

Properties of Limits

Let and be real numbers, let be a positive integer, and let and be functions with the following limits and . Then

1. Scalar multiple:

2. Sum or difference:

3. Product:

4. Quotient:

5. Power:

6. Radical:

Techniques for Evaluating Limits

1. Direct Substitution: Direct Substitution can be used to find the limit of a Polynomial Function.

Example: Evaluate Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\to 3}x^{2}+2x-1=(3)^{2}+2(3)-1=14}

2. Dividing Out Technique: When direct substitution fails and numerator or/and denominator can be factored.

Example: Evaluate . 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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}}} . Now we can use direct substitution to get the answer

One-Sided Limits and Unbounded Function

 when a function approaches a different value from the left of  than it approaches from the right of , 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
 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\to c^{-}}f(x)=L}
 and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\to c^{+}}f(x)=K}

One-sided Limit is related to unbounded function.

In some case, the limit of can be increase/decrease without bound as approaches . We can write Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\to c}f(x)=\pm \infty }

Now, consider . By direct substitution, it is of the form Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {\text{constant}}{0}}} , so the answer will be either or . In order to find the limit, we must consider the limit from both side (Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\to 1^{-}}} and ).

When Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x\to 1^{-}} , so , hence Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x-1<0} . Therefore, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\to 1^{-}}{\frac {-2}{x-1}}={\frac {\text{negative}}{\text{negative}}}=\infty }

When Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x\to 1^{+}} , so Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x>1} , hence . Therefore,



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