Question

The function \[f\left( x \right) = \dfrac{{x(x - 2)}}{{x - 1}}\] is continuous at \[x = 1\]
A) True
B) False

Answer 1

Hint: For determining the continuity of a fraction \[f\left( x \right) = \dfrac{{x(x - 2)}}{{x - 1}}\], first determine the domain of the function , and the function is discontinuous at those points at which any asymptote exists .

Complete step-by-step answer:
Equation is given here
\[f\left( x \right) = \dfrac{{x(x - 2)}}{{x - 1}}\]
Equating the numerator to zero will give the values of the \[x\]for which the function is equal to zero .
\[x(x - 2) = 0\]
\[x = 0;x = 2\] are the two zeroes of the function
Now equating the denominator with zero will give the values of \[x\]for which the denominator is equal to zero (0) . That means in the numerator we are getting zero , there are the points at which the function is not defined or there exists an asymptote of the function.
\[(x - 1) = 0\]
\[x = 1\] is the equation of the vertical asymptote of the function.This means at \[x = 1\] the function is not continuous .
Also the domain is \[x \in ( - \infty ,1) \cup (1,\infty )\]; means the function is discontinuous at \[x = 1\].

Hence the above statement is False.

Note: In the problems in which continuity or discontinuity involved first start with finding the domain of the function , then check for asymptotes an asymptote is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.
All polynomials ,Trigonometric functions, exponential & logarithmic functions are continuous in their domains.