Question 11.3.1: Determining Whether a Function Is Continuous at a Number Det...
Determining Whether a Function Is Continuous at a Number
Determine whether the function
f(x)=\frac{2 x+1}{2 x^2-x-1}
is continuous: a. at 2; b. at 1.
According to the definition, three conditions must be satisfied to have continuity at a.
a. To determine whether f(x)=\frac{2 x+1}{2 x^2-x-1} is continuous at 2 , we check the conditions for continuity with a = 2.
Condition 1 f is defined at a. Is f(2) defined?
f(2)=\frac{2 \cdot 2+1}{2 \cdot 2^2-2-1}=\frac{4+1}{8-2-1}=\frac{5}{5}=1
Because f(2) is a real number, 1, f(2) is defined.
Condition 2 \underset{x\rightarrow a}{\lim} f(x) exists. Does \underset{x\rightarrow 2}{\lim} f(x) exist?
\underset{x\rightarrow 2}{\lim} f(x)=\underset{x\rightarrow 2}{\lim} \frac{2 x+1}{2 x^2-x-1}=\frac{\underset{x\rightarrow 2}{\lim}(2 x+1)}{\underset{x\rightarrow 2}{\lim}\left(2 x^2-x-1\right)}
=\frac{2 \cdot 2+1}{2 \cdot 2^2-2-1}=\frac{4+1}{8-2-1}=\frac{5}{5}=1
Using properties of limits, we see that \underset{x\rightarrow 2}{\lim} f(x) exists.
Condition 3 \underset{x\rightarrow a} {\lim} f(x)=f(a) Does \underset{x\rightarrow 2}{\lim} f(x)=f(2) ? We found that \underset{x\rightarrow 2}{\lim} f(x)=1 and f(2) = 1. Thus, as x gets closer to 2 , the corresponding values of f(x) get closer to the function value at 2: \underset{x\rightarrow 2}{\lim} f(x)=f(2).
Because the three conditions are satisfied, we conclude that f is continuous at 2.
b. To determine whether f(x)=\frac{2 x+1}{2 x^2-x-1} is continuous at 1, we check the conditions for continuity with a = 1.
Condition 1 f is defined at a. Is f(1) defined? Factor the denominator of the function’s equation:
Because division by zero is undefined, the domain of f is \left\{x \mid x \neq 1, x \neq-\frac{1}{2}\right\}.
Thus, f is not defined at 1.
Because one of the three conditions is not satisfied, we conclude that f is not continuous at 1. Equivalently, we can say that f is discontinuous at 1.