The value of x for which the function \(\rm f(x)= \frac{x^2-5x-6}{x^2 +5x-6}\) is not continuous are ?

Concept:

For a function say f, \(\mathop {\lim }\limits_{x \to a} f(x)\) exists

\( ⇒ \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l = \;\mathop {\lim }\limits_{x \to a} f(x)\), where l is a finite value.

Any function say f is said to be continuous at point say a if and only if \(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\), where l is a finite value.

Calculation:

Given: \(\rm f(x)= \frac{x^2-5x-6}{x^2 +5x-6}\)

Here, we have to find the value of x for which f(x) is not continuous.

So, if any function is not continuous at x = a then \(\mathop {\lim }\limits_{x \to a} f(x) = l \neq f\left( a \right)\)

So, for the function f(x) if denominator is 0 at x = a then we can say that f(a) is infinite and limit cannot exist.

Let's find the value of x for which the denominator of f(x) is 0.

⇒ x² + 5x - 6 = 0

⇒ (x + 6) (x - 1) = 0.

⇒ x = -6, 1.

Hence, option 3 is correct.