7. $$ lim _ { x \rightarrow - 1 } \frac { x + 1 } { x ^ { 2 } - 2 x - 3 } $$
7. $$ lim _ { x \rightarrow - 1 } \frac { x + 1 } { x ^ { 2 } - 2 x - 3 } $$
Answer
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\rightarrow - 1}\frac{x + 1}{(x - 3)(x + 1)}).
Step2: Simplify the function
Since (x\neq - 1) (when taking the limit as (x\rightarrow - 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\rightarrow - 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})