Find the continuity of f ( x ) = x 2 - 16 x + 4 , x - 4 at x = k , where k is any positive value:

Question

Find the continuity of $f (x) = \frac{x^{2} - 16}{x + 4}, x \neq - 4$ at $x = k$ , where $k$ is any positive value:

Smart Achievers · Accepted Answer

(2)

Ans. $f (x)$ is continuous at $x = k$

Explanation:

The given function is, $f (x) = \frac{x^{2} - 16}{x + 4}$

At $x = k$ ,

$f (k) = \frac{k^{2} - 16}{k + 4} = \frac{(k + 4) (k - 4)}{k + 4} = k - 4$

$\lim_{x \to k} f (x) = \lim_{x \to k} (\frac{x^{2} - 16}{x + 4}) = \frac{(k + 4) (k - 4)}{k + 4} = k - 4$

$∴$ $\lim_{x \to k} f (x) = f (k)$

$∴$ $f (x) is continuous at x = k$

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