Question
If the statement is alwaystrue, explain why. Ifnot, give a counterexample. If f is apolynomial, then, as x approaches0, theright-hand limit exists and is equal
If the statement is alwaystrue, explain why. Ifnot, give a counterexample.
If f is apolynomial, then, as x approaches0, theright-hand limit exists and is equal to theleft-hand limit.
Choose the correct answer below.
A.
The statement is always true. Iff(x) is apolynomial, then ModifyingBelow lim With x right arrow c f left parenthesis x right parenthesis
limxc
f(x)equals
=c. Because ModifyingBelow lim With x right arrow c f left parenthesis x right parenthesis
limxc
f(x)equals
=L if and only if ModifyingBelow lim With x right arrow c Superscript minus f left parenthesis x right parenthesis
limxc
f(x)equals
=ModifyingBelow lim With x right arrow c Superscript plus f left parenthesis x right parenthesis
limxc+
f(x)equals
=L, it follows by letting cequals
=0 that, as x approaches0, theright-hand limit exists and is equal to theleft-hand limit.
B.
The statement is always true. Iff(x) is apolynomial, then ModifyingBelow lim With x right arrow c f left parenthesis x right parenthesis
limxc
f(x)equals
=f(c). Because ModifyingBelow lim With x right arrow c f left parenthesis x right parenthesis
limxc
f(x)equals
=L if and only if ModifyingBelow lim With x right arrow c Superscript minus f left parenthesis x right parenthesis
limxc
f(x)equals
=ModifyingBelow lim With x right arrow c Superscript plus f left parenthesis x right parenthesis
limxc+
f(x)equals
=L, it follows by letting cequals
=0 that, as x approaches0, theright-hand limit exists and is equal to theleft-hand limit.
C.
The statement is not always true. Forexample, ModifyingBelow lim With x right arrow 0 Superscript minus left parenthesis x minus 1 right parenthesis
limx0
(x1)equals
=minus
1 but ModifyingBelow lim With x right arrow 0 Superscript plus left parenthesis x minus 1 right parenthesis
limx0+
(x1)equals
=1.
D.
The statement is not always true. Forexample, ModifyingBelow lim With x right arrow 0 Superscript minus left parenthesis x squared minus 3 x plus 1 right parenthesis
limx0
x23x+1 does not exist.
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