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valkas [14]
2 years ago
8

Help 13 points!!!!!!!!

Mathematics
1 answer:
DIA [1.3K]2 years ago
3 0

Replace x with a^2:

7(a^2)^2 - 8

Multiply the exponents:

7a^4 - 8

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Step-by-step explanation:

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For what value of constant c is the function k(x) continuous at x = 0 if k =
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The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

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Learn more about the limit of a function x here:

brainly.com/question/8131777

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(suprisingly easy) 6x3x4=72 in.

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