The limit on the left cannot be evaluated by direct substitution because if 2 is substituted in, then you end up dividing by zero.The limit on the right can be evaluated using direct substitution beca...The limit on the left cannot be evaluated by direct substitution because if 2 is substituted in, then you end up dividing by zero.The limit on the right can be evaluated using direct substitution because the hole exists at x=2 not x=3. Substitution is a method of determining limits where the value that x is approaching is substituted into the function and the result is evaluated. A limit is the value that the output of a function approaches as the input of the function approaches a given value.
Hence, the function has a zero at x=2, there is a hole in the graph at x=−2, the domain is (−∞,−2)∪(−2,4)∪(4,+∞) \nonumber, and the y-intercept is at (0,12). \[ \lim_{x \to ∞} \frac{x^2−4}{x^2−2x...Hence, the function has a zero at x=2, there is a hole in the graph at x=−2, the domain is (−∞,−2)∪(−2,4)∪(4,+∞) \nonumber, and the y-intercept is at (0,12). \lim_{x \to ∞} \frac{x^2−4}{x^2−2x−8}= \lim_{x \to ∞} \frac{ \frac{x^2}{x^2}−\frac{4}{x^2}} {\frac{x^2}{x^2}−\frac{2x}{x^2}−\frac{8}{x^2}}=\lim_{x \to ∞} \frac{1−\frac{4}{x^2}}{1−\frac{2}{x}−\frac{8}{x^2}}=1 \nonumber. The function can be factored f(x)=x^3+2x^2−x−2=x^2(x+2)−1(x+2)=(x^2−1)(x+2)=(x−1)(x+1)(x+2) \nonumber
