5 c) Examine the following functions for continuity.

f (x) = \frac{x ^2-25}{x+5}, x \neq -5

Answers (1)

Given function is
f(x ) = \frac{x^2-25}{x+5}
For every real number  k , k \neq -5
We gwt,
f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = f(k)
Hence, function  f(x ) = \frac{x^2-25}{x+5} continuous for every real value of x , x \neq -5

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