Answer:
As x→[infinity], f(x)→ 1
As x→−[infinity], f(x)→ 1
As x→−5+, f(x)→ -∞
As x→−5−, f(x)→ ∞
Explanation:
x→∞ f(x)=(x − 8)/(x + 5) ∞/∞ ==> the indeterminate form
x→∞ f(x)=(x − 8)/x / (x + 5)/x ==> divide each side by x
x→∞ f(x)=(x/x − 8/x) / (x/x + 5/x) ==> now each term is being divided by x
x→∞ f(x)=(1 − 8/x) / (1 + 5/x) ==> simplify
f(x)=(1 − 8/∞) / (1 + 5/∞) ==> x = ∞
f(x)=(1−0) / (1+0) ==> 5/∞ and 8/∞ are basically 0 since any number
divided by an extremely large number like ∞
results in a number that is nearly 0.
f(x)=1/1 ==> simplify
f(x) → 1
f(x)=(1 − 8/x) / (1 + 5/x) ==> now use this for -∞:
x→-∞ f(x)=(1 − 8/x) / (1 + 5/x)
f(x)=(1 − 8/(-∞)) / (1 + 5/(-∞)) ==> x = -∞
f(x)=(1−0) / (1+0) ==> 5/-∞ and 8/-∞ are also basically 0
f(x) → 1
x→−5+ ==> x + 5 = 0+ where 0+ is a number slightly greater than 0 but is a
very small number
x→−5+ f(x)=(x − 8)/(x + 5) =
x→−5+ f(x)=(x − 8)/0+ =
f(x)=(-5 − 8)/0+ =
f(x) = -13/0+
f(x) → -∞ ==> any number divided by a number that is extremely small is ∞.
Now solve for x→−5−, f(x)→:
x→−5- ==> x + 5 = 0- where 0- is a number slightly less than 0 and is
a very small negative number
x→−5- f(x)=(x − 8)/(x + 5) =
x→−5- f(x)=(x − 8)/0- =
f(x)=(-5 − 8)/0+ =
f(x) = -13/0-
f(x) = 13/0+ ==> cancel the negative signs
f(x) → ∞
Hence:
x→∞ f(x) ==> f(x) → 1
x→-∞ f(x) ==> f(x) → 1
x→-5+ f(x) ==> f(x) → -∞
x→-5- f(x) ==> f(x) → ∞