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Consider f(x)=x−8/x+5. Complete the following statements.

As x→[infinity], f(x)→

As x→−[infinity], f(x)→

As x→−5+, f(x)→

As x→−5−, f(x)→




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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) → ∞

User Larry Watanabe
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