Question

How do you find the limit of $\dfrac{{{x^2} - x - 6}}{{{x^2} - 5x + 6}}$ as $x$ approaches to 0?

Answer 1

To find the limit of the given equation, as the equation contains $x$ and $x$ in which $x$ approaches 0 implies that the denominator approaches 6 and the numerator approaches to -6. Hence, this form is not indeterminate i.e., the limit cannot be determined by individual functions.

Complete step by step solution:
Let us write the given input
$\dfrac{{{x^2} - x - 6}}{{{x^2} - 5x + 6}}$
As $x$ approaches to 0, as
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - x - 6}}{{{x^2} - 5x + 6}}$
We get
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - 0 - 6}}{{{x^2} - 0 + 6}}$
Therefore,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - x - 6}}{{{x^2} - 5x + 6}} = \dfrac{{ - 6}}{6} = - 1$

As $x \to 0$, the denominator approaches 6. The form is not indeterminate.

Additional information:
Here are some of the properties to find the limit functions:
Sum Rule: This rule states that the limit of the sum of two functions is equal to the sum of their limits.
Constant Function Rule: The limit of a constant function is the constant.
Constant Multiple Rule: The limit of a constant times a function is equal to the product of the constant and the limit of the function.
Product Rule: This rule says that the limit of the product of two functions is the product of their limits (if they exist).
Quotient Rule: The limit of quotient of two functions is the quotient of their limits, provided that the limit in the denominator function is not zero.

Note:
For a limit approaching the given value, the original functions must be differentiable on either side of value, but not necessarily at the value given. The limit of a quotient is equal to the quotient of the limits. The limit of a constant function is equal to the constant. The limit of a linear function is equal to the number x is approaching.