Question

(a) What is wrong with the following equation?

\(\frac{{{x^2} + x - 6}}{{x - 2}} = x + 3\)

(b) In view of part (a), explain why the equation

\(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x - 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \,x + 3\)

is correct

Short answer

1. This equation does not hold for \(x = 2\).
2. As \(x\) approaches 2, both sides of the equation become equal. The reason is on calculating \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\), the point \(x = 2\) is never considered.

Step-by-step solution

Analyze the given equation for any discontinuity

The first equation is\(\frac{{{x^2} + x - 6}}{{x - 2}} = x + 3\).

In this equation, the left side has a rational polynomial, which is not defined for\(x = 2\), or its denominator becomes 0 for\(x = 2\).

Whereas the right part is a monomial, which is defined for all\(x\).

Thus, this equation does not hold for \(x = 2\).

Use the definition of limit

The equation does not hold for\(x = 2\), but as\(x\)approaches 2, both sides of the equation become equal.

The reason is on calculating \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\), the point \(x = 2\) is never considered.