Paul H. answered • 09/11/22
Proficiency in Algebra is key to success in calculus and later math.
Note that rational functions are continuous except where the denominator would become zero.
g(x) = (x^2 − 16) / (x^3 − 64)
The denominator x^3 - 64 has its only real zero at x=4. (Determine this by setting to zero and solving by factoring or by looking at its graph.)
Part c) first. Since g(x) is continuous at x=-4 (which is not x=4, the pt of discontinuity), it is valid to get the limit as g(-4) = ( (-4)^2 - 16) / ((-4)^3 -64) = (16 - 16) / (-64 - 64) = 0 / -128 = 0 // (just used substitution)
Part a) next. Since g(x) is NOT continuous at x=4, substitution is not valid. Note that substitution of 4 would give us 0/0 , that is, both numerator and denominator would be zero. That is a sure sign that we have to use algebra to simplify and continue. Since x=4 is a zero of each of the numerator and the denominator, it is guaranteed that x-4 is a factor of each! Do that factoring:
g(x) = (x^2 − 16) / (x^3 − 64) = [ (x-4) (x+4) ] / [ (x-4) (x^2 +4x +16) ] (use long division to get that last factor, if you forgot how to factor the difference of two cubes)
Simplify by cancelling the x-4 factors. (Legal, as long as not cancelling zeros! but x cannot =4)
Now, h(x) = (x+4) / (x^2 +4x +16) is the same function as g(x), provided x is not 4.
But since a limit looks at behavior CLOSE to an x value, not AT that x value, we can see the limit of h(x) as x nears 4. And h(x) IS CONTINUOUS at =4 (denominator not zero there)!, so now the substitution into it IS VALID. h(4) = (4+4) / (4^2 +4*4 +16) = 8/48 = 1/6 // So that is also the limit for g(x) as x-> 4. //
Part b) now. New ball game, since ∞ is not a number, just a shortcut for saying 'things get big'.
For you or me, 1 Thousand $ (10^3=1,000) is a lot, but not for our federal government. It pales compared to 1 Million $ (10^6=1,000,000); which itself pales compared to 1 Billion dollars (10^9=1,000,000,000). In a budget of Billions, then Thousands or even Millions are not significant in comparison. Bigger powers of 10 make smaller powers irrelevant. In our math, it will be bigger powers of the variable making the smaller ones irrelevant, IN THE LONG RUN.
for g(x) = (x^2 − 16) / (x^3 − 64) , suppose x were to be 1000. Then the 16 is irrelevant in the numerator and same for the 64 in the denominator.
(1000^2) / (1000^3) = .001000000000 while (1000^2 − 16) / (1000^3 − 64) = .000999984064 Close! Difference in 8th decimal place. Make x bigger, and results will be even closer.
For BIG x, g(x) is very close to x^2 / x^3 = 1 / x. And what happens to that as x -> ∞ ?
Since x gets very large Positive, 1/x gets very small positive, goes to zero (but on the positive side).
Fancy says lim x→∞ of g(x) is 0+. (But if x→-∞, 1/x →0- (that is, values go to zero but are negative).
Note. Some teachers don't like the - or + on the 0. Some teachers don't like this "dominant term" method, where you ignore terms added or subtracted that are more insignificant -- they prefer to mess down in the weeds with a lot more algebra -- make your teacher happy.
