Question

Determine the infinite limit.

\(\mathop {\lim }\limits_{x \to {3^ - }} \,\frac{{{x^2} + 4x}}{{{x^2} - 2x - 3}}\)

Short answer

The limit tends to negative infinity.

Step-by-step solution

Analyze the given function at the given point

Consider the limit\(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} + 4x}}{{{x^2} - 2x - 3}}\).

Factorize the denominator of the function as shown below:

\(\begin{aligned}{x^2} - 2x - 3 &= {x^2} - 3x + x - 3\\ &= x\left( {x - 3} \right) + 1\left( {x - 3} \right)\\ &= \left( {x - 3} \right)\left( {x + 1} \right)\end{aligned}\)

The given function’sdenominator can be factored as\(\left( {x - 3} \right)\left( {x + 1} \right)\), which approaches 0 from left as\(x\)approaches 3 from left.

The numerator is a positive value approaching 21 from left as\(x\)approaches 3 from left.

Since the denominator tends to zero as \(x\) approaches 3 from left, so the function becomes discontinuous as \(x \to {3^ - }\).

Estimate the limit

The numerator tends to a positive value, and the denominator of the function tends to 0 fromleft as\(x\)approaches3 from left.

Thus,the final limitmustapproachtonegative infinityas shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to {3^ - }} \frac{{{x^2} + 4x}}{{{x^2} - 2x - 3}} &= \frac{{{{21}^ - }}}{{{0^ - }}}\\ &= - \infty \end{aligned}\)

Thus, the limit tends to negative infinity.