Question

How do you find the x values at which \[f\left( x \right) = \dfrac{x}{{{x^2} - 1}}\] is not continuous, which of the discontinuities are removable ?

Answer 1

Hint:If a function is not continuous at a point in its domain, it implies that the discontinuity is present in the function and a function is continuous if the point is defined and it exists within the given range and to find the x values of the function, as it is a rational function, we know that these are continuous on their domains except \[ \pm 1\]. Hence by this we can get to know at what value the function is discontinuous.

Complete step by step answer:
The given equation is \[f\left( x \right) = \dfrac{x}{{{x^2} - 1}}\]. This is a rational function.Rational functions are continuous on their domains. The domain of this function $f$ is all reals except ±1. So, $f$ is discontinuous at ±1. A discontinuity at a is removable if \[\mathop {\lim }\limits_{x \to a} f\left( x \right)\] exists. But the limit of function $f$ fails to exist for both 1 and −1. So, neither discontinuity is removable.

Additional information:
Continuity Definition: A function is said to be continuous in a given interval if there is no break in the graph of the function in the entire interval range. Assume that “f” be a real function on a subset of the real numbers and “c” be a point in the domain of f. Then f is continuous at c if
\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\]

Discontinuity Definition: The function “f” will be discontinuous at x = a in any of the following cases:
If f (a) is not defined, \[\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)\] and \[\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right)\] exists but are not equal and if \[\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)\] and \[\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right)\] exists are equal but not equal to f (a).

Note: As we know that rational function can be written as the ratio of two polynomial functions, where the polynomial in the denominator is not equal to zero and these are continuous on their domains except \[ \pm 1\]. Hence based on this we can find the values of x of the function given.