Question

7.

$$ lim _ { x ightarrow - 1 } \frac { x + 1 } { x ^ { 2 } - 2 x - 3 } $$

Explanation:

Step1: Factor the denominator

Factor \(x^{2}-2x - 3\) using the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\), where \(x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For \(a = 1\), \(b=-2\), \(c=-3\), \(x=\frac{2\pm\sqrt{4+12}}{2}=\frac{2\pm4}{2}\), so \(x_1 = 3\), \(x_2=-1\). Then \(x^{2}-2x - 3=(x - 3)(x + 1)\).
The limit becomes \(\lim_{x
ightarrow - 1}\frac{x + 1}{(x - 3)(x + 1)}\).

Step2: Simplify the function

Since \(x
eq - 1\) (when taking the limit as \(x
ightarrow - 1\), \(x\) approaches \(-1\) but is not equal to \(-1\)), we can cancel out the common factor \((x + 1)\). The function simplifies to \(\lim_{x
ightarrow - 1}\frac{1}{x - 3}\).

Step3: Substitute \(x=-1\)

Substitute \(x=-1\) into \(\frac{1}{x - 3}\), we get \(\frac{1}{-1-3}\).

Answer:

\(-\frac{1}{4}\)