Quiz:Limit

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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

1 Which of these is the correct interpretation of $\lim _{x\to c}f(x)=L$ 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 $|f(x)-L|<\varepsilon$.