Answer: lim as x → -5 of f(x) and g(x) = 300
Domain of f(x) and g(x) is All Real Numbers
f(-5) = 200
g(-5) = 300
Explanation:
The limit of f(x) as x approaches -5:
![f(x) =(4x^3+500)/(x+5)\\\\\\.\qquad =(4(x^3+125))/(x+5)\\\\\\.\qquad =(4(x^3+5^3))/(x+5)\qquad \text{we can factor the cubic}\\\\\\.\qquad =(4(x+5)(x^2-5x+25))/(x+5)\\\\\\.\qquad =4(x^2-5x+25)\\\\\\\text{as x approaches -5, f(x) = }4[(-5)^2-5(-5)+25]\\\\.\qquad \qquad \qquad \qquad \qquad =4(25 + 25 + 25)\\\\.\qquad \qquad \qquad \qquad \qquad =4(75)\\\\.\qquad \qquad \qquad \qquad \qquad =300](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tccabjh4ot8gu9nvo5cfaagdxcux4utmo8.png)
The limit of g(x) as x approaches -5:
g(x) = 4x² - 20x + 100
= 4(-5)² - 20(-5) + 100
= 100 + 100 + 100
= 300
There is a restriction at x = -5 for f(x), however, that discontinuity has been filled with the 200 at x = -5. So the domain is ALL REAL NUMBERs.
There are no restrictions on x for g(x) so the domain is ALL REAL NUMBERs.