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.

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.

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.

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.

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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