Quiz:Limit

ORIGINAL FULL PAGE: Limit
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Motivation

Two key ideas

1 Suppose $f$ is a function defined on all of $\mathbb{R}$. We find that $\underset{x\to 2}{lim}f(x)=5$. Suppose $g$ is another function defined on $\mathbb{R}$ that differs from $f$ at one point $p$, i.e., $f(x)=g(x)$ for all $x\ne p$, but $f(p)\ne g(p)$. Which of the following is true?

	Whatever the value of $p$, $\underset{x\to 2}{lim}g(x)=5$
	If $p\ne 2$, then $\underset{x\to 2}{lim}g(x)=5$, but if $p=2$, we cannot say anything about the limit.
	If $p\ne 2$, then $\underset{x\to 2}{lim}g(x)=5$, but if $p=2$, then $\underset{x\to 2}{lim}g(x)\ne 5$
	$\underset{x\to 2}{lim}g(x)=5$ unless $p$ is very close to (but still not equal to) 2 . In case $p$ is very close to (but still not equal to) 2, we cannot say anything about the limit.
	$\underset{x\to 2}{lim}g(x)=5$ unless $p$ is very close to (but still not equal to) 2 . In case $p$ is very close to (but still not equal to) 2, the limit is definitely not equal to 5.

2 Which of the following is the best verbal explanation of why the limit $\underset{x\to 0}{lim}\sin (1/x)$ does not exist? As a sanity check for your answer option, keep in mind that $\underset{x\to 0}{lim}x\sin (1/x)=0$, so your answer option should not predict that $\underset{x\to 0}{lim}x\sin (1/x)$ does not exist.

	When $x=0$, $1/x$ is undefined, so $\sin (1/x)$ does not make sense at the point 0.
	The function $\sin (1/x)$ oscillates between positive and negative values for $x$ arbitrarily close to zero.
	$\sin (1/x)$ cannot be trapped in any interval of width less than two for $x$ in an arbitrarily small neighborhood of zero.

Definition for finite limit for finite function of one variable

Two-sided limit

1 Suppose $c,L\in \mathbb{R}$ and $f$ is a function defined on a subset of $\mathbb{R}$. Which of these is the correct interpretation of $\underset{x\to c}{lim}f(x)=L$ in terms of the definition of limit?

	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 $f$ is a function defined on some subset of $\mathbb{R}$. Suppose $c$ and $L$ are real numbers. If $\underset{x\to c}{lim}f(x)=L$, what can we say about $f(c)$?

	$f(c)$ exists and is equal to $L$.
	$f(c)$ does not exist.
	$f(c)$ may or may not exist, but if it exists, it must equal $L$.
	$f(c)$ must exist, but it need not be equal to $L$.
	$f(c)$ may or may not exist, and even if it does exist, it may or may not be equal to $L$.

Left hand limit and right hand limit

1 Which of these is the correct interpretation of the left hand limit $\underset{x\to c^{-}}{lim}f(x)=L$ in terms of the definition of limit?

	For every $\epsilon >0$, there exists $\delta >0$ such that for all $x\in \mathbb{R}$ satisfying $0<c-x<\delta$, we have $0<L-f(x)<\epsilon$.
	For every $\epsilon >0$, there exists $\delta >0$ such that for all $x\in \mathbb{R}$ satisfying $0<x-c<\delta$, we have $0<f(x)-L<\epsilon$.
	For every $\epsilon >0$, there exists $\delta >0$ such that for all $x\in \mathbb{R}$ satisfying $0<x-c<\delta$, we have $0<L-f(x)<\epsilon$.
	For every $\epsilon >0$, there exists $\delta >0$ such that for all $x\in \mathbb{R}$ satisfying $0<|x-c|<\delta$, we have $0<L-f(x)<\epsilon$.
	For every $\epsilon >0$, there exists $\delta >0$ such that for all $x\in \mathbb{R}$ satisfying $0<c-x<\delta$, we have $|f(x)-L|<\epsilon$.

2 Suppose the domain of a function $f$ is a closed bounded interval, i.e., an interval of the form $[a,b]$ for real numbers $a,b$. Which of the following definitely do not make sense?

	The left hand limit at the left endpoint and the right hand limit at the right endpoint.
	The left hand limit at the right endpoint and the right hand limit at the left endpoint.
	The left hand limit and the right hand limit at any interior point.
	The two-sided limit at any interior point.
	The left hand limit at any point other than the left endpoint and the right hand limit at any point other than the right endpoint.

Definition of finite limit for function of one variable in terms of a game

In the usual $\epsilon -\delta$ definition of limit for a given limit $\underset{x\to c}{lim}f(x)=L$, 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.

Non-existence of limit

Suppose $f$ is a function defined on a subset of $\mathbb{R}$. $c$ is a real number such that $(c-t,c)\cup (c,c+t)$ is in the domain of $f$ for some $t>0$. Identify the correct interpretation of the statement "$\underset{x\to c}{lim}f(x)$ does not exist" among the choices below.

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