1. Limit [x -> 4] f(x), where f(x) = (sqrt(x) - 2)/(x - 4)
We can take the "difference of two squares" for (x - 4) to give (sqrt(x) - 2)(sqrt(x) + 2)
Which means that we can cancel the term (sqrt(x) - 2) in f(x) and write it as,
f(x) = 1/(sqrt(x) + 2))
As x - 4, f(x=4) tends to 1/(2 + 2) = 1/4
Answer: LImit = 1/4
2. Limit [x -> 0] f(x), where f(x) = ((x - 5)^2 - 25)/x
Again, using "difference of two squares" f(x) = ((x - 5) - 5))((x - 5) + 5)/x = (x - 10)
As x -> 0, f(x=0) -> -10
Answer: Limit = -10