Question

Examine the following functions for continuity.
(A) f(x) = x-5
(B) f(x) =$\frac {1}{x-5}$, x≠5
(C) f(x) = $\frac {x^2-25}{x+5}$, x≠-5
(D) f(x) = |x-5|

Answer 1

(A) The given function is f(x) = x-5
It is evident that f is defined at every real number k and its value at k is k−5.
It is also observed that,
$\lim\limits_{x \to k}$ f(x) = $\lim\limits_{x \to k}$ (x-5) = k-5 = f(k)
∴$\lim\limits_{x \to k}$ f(x) = f(k)
Hence, f is continuous at every real number and therefore, it is a continuous function.

(B) The given function is f(x) = $\frac {1}{x-5}$, x≠5
For any real number k ≠ 5, we obtain
$\lim\limits_{x \to k}$ f(x) = $\lim\limits_{x \to k}$ $\frac {1}{x-5}$ = $\frac {1}{k-5}$
Also,f(k) = $\frac {1}{k-5}$ (As k≠5)
∴$\lim\limits_{x \to k}$ f(x) = f(k)
Hence,f is continuous at every point in the domain of f and therefore,it is a continuous function.

(C) The given function is f(x)=$\frac {x^2-25}{x+5}$, x≠-5
For any real number c≠−5,we obtain
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ $\frac {x^2-25}{x+5}$ = $\lim\limits_{x \to c}$ $\frac {(x-5)(x+5)}{x+5}$ = $\lim\limits_{x \to c}$ (x-5) = (c-5)
Also ,f(c) = $\frac {(c-5)(c+5)}{c+5}$ = c-5 (As c≠-5)
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Hence,f is continuous at every point in the domain of f and therefore,it is a continuous function.

(D) The given function is

$f(x)= |x-5|=\begin{cases} 5-x & \quad \text{if } n \text{<5}\\\ x-5 & \quad \text{if } x {\geq 5} \end{cases}$

This function f is defined at all points of the real line.
Let c be a point on a real line.
Then, c<5 or c=5 or c>5
Case I: c<5
Then, f(c) = 5−c
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (5-x) = 5-c
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore,f is continuous at all real numbers less than 5.

Case II: c = 5
Then, f(c) = f(5) = 5−5 = 0
$\lim\limits_{x \to 5^-}$ f(x) = $\lim\limits_{x \to 5}$ (5-5)=0
$\lim\limits_{x \to 5^+}$ f(x) = $\lim\limits_{x \to 5}$ (x-5) = 0
∴$\lim\limits_{x \to c^-}$ f(x) = $\lim\limits_{x \to c^+}$ f(x) = f(c)
Therefore, f is continuous at x = 5

Case III: c>5
Then, f(c) =f(5) = c-5
$\lim\limits_{x \to c}$ f(x) =$\lim\limits_{x \to c}$ (x-5) = c-5
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore,f is continuous at all real numbers greater than 5.

Hence,f is continuous at every real number and therefore, it is a continuous function.