Quiz:Limit

ORIGINAL FULL PAGE: Limit
STUDY THE TOPIC AT MULTIPLE LEVELS:
ALSO CHECK OUT: Quiz (multiple choice questions to test your understanding) |Page with videos on the topic, both embedded and linked to

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 $\lim _{x\to c}f(x)=L$ 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 $\lim _{x\to c}f(x)=L$, 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 $\lim _{x\to c^{-}}f(x)=L$ in terms of the definition of limit?

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

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.