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.