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Questions and Answers
When can the limit of a continuous function be found using direct substitution?
When can the limit of a continuous function be found using direct substitution?
- When the function is undefined at that point.
- Only for polynomial functions.
- Always.
- When the function is continuous at that point. (correct)
Direct substitution can always be used to evaluate limits of rational functions.
Direct substitution can always be used to evaluate limits of rational functions.
False (B)
What is the primary reason why factorization is used when evaluating limits in indeterminate forms?
What is the primary reason why factorization is used when evaluating limits in indeterminate forms?
to cancel common factors
The expression (\mathop {\lim }\limits_{x o 1} rac{{{x^2} - 1}}{{x - 1}}) is indeterminate because the factor _______ makes the limit of the form ( rac{0}{0}).
The expression (\mathop {\lim }\limits_{x o 1} rac{{{x^2} - 1}}{{x - 1}}) is indeterminate because the factor _______ makes the limit of the form ( rac{0}{0}).
What is the limit of the expression (\mathop {\lim }\limits_{x o 1} rac{{{x^m} - 1}}{{x - 1}})?
What is the limit of the expression (\mathop {\lim }\limits_{x o 1} rac{{{x^m} - 1}}{{x - 1}})?
Rationalization of an indeterminate expression always leads to a determinate limit.
Rationalization of an indeterminate expression always leads to a determinate limit.
What indeterminate form does the expression (\mathop {\lim }\limits_{x o \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} }
ight)) initially represent?
What indeterminate form does the expression (\mathop {\lim }\limits_{x o \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } ight)) initially represent?
In the example (\mathop {\lim }\limits_{x o rac{\pi }{2}} rac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}}), the substitution (x = rac{\pi }{2} + h) is used so that as (x o rac{\pi }{2}), (h) approaches _______.
In the example (\mathop {\lim }\limits_{x o rac{\pi }{2}} rac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}}), the substitution (x = rac{\pi }{2} + h) is used so that as (x o rac{\pi }{2}), (h) approaches _______.
What trigonometric identity is used to simplify the expression ( rac{{1 - \cos , h}}{{{h^2}}}) in the limit (\mathop {\lim }\limits_{h o 0} ,, rac{{\sin , h}}{h}. rac{{1 - \cos , h}}{{{h^2}}}. rac{1}{{\cos , h}})?
What trigonometric identity is used to simplify the expression ( rac{{1 - \cos , h}}{{{h^2}}}) in the limit (\mathop {\lim }\limits_{h o 0} ,, rac{{\sin , h}}{h}. rac{{1 - \cos , h}}{{{h^2}}}. rac{1}{{\cos , h}})?
Polynomial limits can always be evaluated by direct substitution because polynomials are continuous everywhere.
Polynomial limits can always be evaluated by direct substitution because polynomials are continuous everywhere.
In the context of evaluating limits, what does 'determinate form' mean?
In the context of evaluating limits, what does 'determinate form' mean?
The method of _______ is used to eliminate radicals in expressions to evaluate limits, especially when dealing with indeterminate forms.
The method of _______ is used to eliminate radicals in expressions to evaluate limits, especially when dealing with indeterminate forms.
Match the method of limit evaluation with the corresponding example:
Match the method of limit evaluation with the corresponding example:
Which of the following is a type of indeterminate form encountered when evaluating limits?
Which of the following is a type of indeterminate form encountered when evaluating limits?
If a function is discontinuous at a point, its limit at that point cannot exist.
If a function is discontinuous at a point, its limit at that point cannot exist.
What algebraic manipulation is typically used after rationalization to simplify the limit expression?
What algebraic manipulation is typically used after rationalization to simplify the limit expression?
When evaluating limits of the form (\mathop {\lim }\limits_{x o \infty } rac{{f(x)}}{{g(x)}}), dividing both the numerator and denominator by the highest power of $x$ is a method to convert it to a _______ form.
When evaluating limits of the form (\mathop {\lim }\limits_{x o \infty } rac{{f(x)}}{{g(x)}}), dividing both the numerator and denominator by the highest power of $x$ is a method to convert it to a _______ form.
Match the limit expression with the appropriate method to simplify it:
Match the limit expression with the appropriate method to simplify it:
In evaluating limits, what does it mean for a form to be 'indeterminate'?
In evaluating limits, what does it mean for a form to be 'indeterminate'?
Substitution is only useful when direct evaluation results in a determinate form.
Substitution is only useful when direct evaluation results in a determinate form.
Flashcards
Continuous Function Limit
Continuous Function Limit
For continuous functions, the limit at a point can be found by direct substitution.
Polynomial Limit Evaluation
Polynomial Limit Evaluation
Polynomial limits can be evaluated by direct substitution.
Indeterminate Form
Indeterminate Form
An indeterminate form arises when a certain factor in an expression makes the limit undefined, often 0/0.
Factorization Method
Factorization Method
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Rationalization Method
Rationalization Method
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Indeterminate Form ∞ - ∞
Indeterminate Form ∞ - ∞
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Trigonometric Substitution
Trigonometric Substitution
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Study Notes
- For continuous functions, limits are found via direct substitution due to their definition.
- Polynomial limits can be evaluated by direct substitution.
- Indeterminate forms in limits occur when a factor in the expression, like (x – 1), makes the limit undefined, such as (\frac{0}{0}).
- Factorization cancels the common factor, simplifying the limit to a determinate form.
Factorization Examples
- (\begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{x - 1}} = 3\end{align})
- This limit resembles (\begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{{x^m} - 1}}{{x - 1}}\end{align}), which equals m.
- (\begin{align}\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{{x^2} - 4}} = \frac{1}{4}\end{align})
- (\begin{align}\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^6} - 1}}{x} = 6\end{align})
Rationalization
- Rationalization transforms an indeterminate expression into a determinate one
- This is achieved when direct substitution results in both numerator and denominator equaling 0
Rationalization Examples
- (\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}}) is of the indeterminate form (\frac{{\rm{0}}}{{\rm{0}}})
- (\begin{align}\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } \right)\end{align}) is of the indeterminate form (\infty - \infty )
Substitution
- Substitution can be used when encountering an indeterminate form of (\frac{0}{0})
- Substituting (x = \frac{\pi }{2} + h) so that as (x \to \frac{\pi }{2},,, h \to 0) allows the limit to be solved
- (\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}} = \frac{1}{{16}})
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