Quiz:Limit and continuity: Difference between revisions

Revision as of 23:06, 20 October 2011

Formal definition of limit and continuity

	For every $\alpha >0$ , there exists $\beta >0$ such that if $0<\|x-c\|<\alpha$ , then $\|f(x)-L\|<\beta$ .
	There exists $\alpha >0$ such that for every $\beta >0$ , and $0<\|x-c\|<\alpha$ , we have $\|f(x)-L\|<\beta$ .
	For every $\alpha >0$ , there exists $\beta >0$ such that if $0<\|x-c\|<\beta$ , then $\|f(x)-L\|<\alpha$ .
	There exists $\alpha >0$ such that for every $\beta >0$ and $0<\|x-c\|<\beta$ , we have $\|f(x)-L\|<\alpha$ .
	None of the above

2 Suppose $f:\mathbb {R} \to \mathbb {R}$ is a function. Which of the following says that $f$ does not have a limit at any point in $\mathbb {R}$ (i.e., there is no point $c\in \mathbb {R}$ for which $\lim _{x\to c}f(x)$ exists)?

	For every $c\in \mathbb {R}$ , there exists $L\in \mathbb {R}$ such that for every $\epsilon >0$ , there exists $\delta >0$ such that for all $x$ satisfying $0<\|x-c\|<\delta$ , we have <math>\|f(x) - L\| \ge
	There exists $c\in \mathbb {R}$ such that for every $L\in \mathbb {R}$ , there exists $\epsilon >0$ such that for every $\delta >0$ , there exists $x$ satisfying $0<\|x-c\|<\delta$ and $\|f(x)-L\|\geq \epsilon$ .
	For every $c\in \mathbb {R}$ and every $L\in \mathbb {R}$ , there exists $\epsilon >0$ such that for every $\delta >0$ , there exists $x$ satisfying $0<\|x-c\|<\delta$ and $\|f(x)-L\|\geq \epsilon$ .
	There exists $c\in \mathbb {R}$ and $L\in \mathbb {R}$ such that for every $\epsilon >0$ , there exists $\delta >0$ such that for all $x$ satisfying $0<\|x-c\|<\delta$ , we have $\|f(x)-L\|\geq \epsilon$ .
	All of the above.

3 In the usual $\epsilon -\delta$ definition of limit for a given limit $\lim _{x\to c}f(x)=L$ , if a given value $\delta >0$ works for a given value $\epsilon >0$ , then which of the following is true?

	Every smaller positive value of $\delta$ works for the same $\epsilon$ . Also, the given value of $\delta$ works for every smaller positive value of $\epsilon$ .
	Every smaller positive value of $\delta$ works for the same $\epsilon$ . Also, the given value of $\delta$ works for every larger value of $\epsilon$ .
	Every larger value of $\delta$ works for the same $\epsilon$ . Also, the given value of $\delta$ works for every smaller positive value of $\epsilon$ .
	Every larger value of $\delta$ works for the same $\epsilon$ . Also, the given value of $\delta$ works for every larger value of $\epsilon$ .
	None of the above statements need always be true.

@@ Line 26: / Line 26: @@
 - Every larger value of <math>\delta</math> works for the same <math>\epsilon</math>. Also, the given value of <math>\delta</math> works for every larger value of <math>\epsilon</math>.
 - None of the above statements need always be true.
+{Which of the following is a correct formulation of the statement <math>\lim_{x \to c} f(x) = L</math>, in a manner that avoids the use of <math>\epsilon</math>s and <math>\delta</math>s?
+- For every open interval centered at <math>c</math>, there is an open interval centered at <math>L</math> such that the image under <math>f</math> of the open interval centered at <math>c</math> (excluding the point <math>c</math> itself) is contained in the open interval centered at <math>L</math>.
+- For every open interval centered at <math>c</math>, there is an open interval centered at <math>L</math> such that the image under <math>f</math> of the open interval centered at <math>c</math> (excluding the point <math>c</math> itself) contains the open interval centered at <math>L</math>.
++ For every open interval centered at <math>L</math>, there is an open interval centered at <math>c</math> such that the image under <math>f</math> of the open interval centered at <math>c</math> (excluding the point <math>c</math> itself) is contained in the open interval centered at <math>L</math>.
+- For every open interval centered at <math>L</math>, there is an open interval centered at <math>c</math> such that the image under <math>f</math> of the open interval centered at <math>c</math> (excluding the point <math>c</math> itself) contains the open interval centered at <math>L</math>.
+- None of the above.
 </quiz>

	For every $\alpha >0$ , there exists $\beta >0$ such that if $0<\|x-c\|<\alpha$ , then $\|f(x)-L\|<\beta$ .
	There exists $\alpha >0$ such that for every $\beta >0$ , and $0<\|x-c\|<\alpha$ , we have $\|f(x)-L\|<\beta$ .
	For every $\alpha >0$ , there exists $\beta >0$ such that if $0<\|x-c\|<\beta$ , then $\|f(x)-L\|<\alpha$ .
	There exists $\alpha >0$ such that for every $\beta >0$ and $0<\|x-c\|<\beta$ , we have $\|f(x)-L\|<\alpha$ .
	None of the above

	For every $c\in \mathbb {R}$ , there exists $L\in \mathbb {R}$ such that for every $\epsilon >0$ , there exists $\delta >0$ such that for all $x$ satisfying $0<\|x-c\|<\delta$ , we have <math>\|f(x) - L\| \ge
	There exists $c\in \mathbb {R}$ such that for every $L\in \mathbb {R}$ , there exists $\epsilon >0$ such that for every $\delta >0$ , there exists $x$ satisfying $0<\|x-c\|<\delta$ and $\|f(x)-L\|\geq \epsilon$ .
	For every $c\in \mathbb {R}$ and every $L\in \mathbb {R}$ , there exists $\epsilon >0$ such that for every $\delta >0$ , there exists $x$ satisfying $0<\|x-c\|<\delta$ and $\|f(x)-L\|\geq \epsilon$ .
	There exists $c\in \mathbb {R}$ and $L\in \mathbb {R}$ such that for every $\epsilon >0$ , there exists $\delta >0$ such that for all $x$ satisfying $0<\|x-c\|<\delta$ , we have $\|f(x)-L\|\geq \epsilon$ .
	All of the above.