- Topic
- Limits
- Difficulty
- Standard
- Suggested time
- 3 min
- Marks
- 2
Limits practice question 19
The graph of the function below has a different horizontal asymptote at each end. Find the one it approaches as : enter its value of .
This question asks about only one end of the graph. What does do at that end?
Treat the two ends separately. As , does not grow: it shrinks towards 0.
So substitute the limit of into the top and the bottom. Dividing by is the move for , not here.
An exact value, or a decimal to three places.
Theory UBC Math 100 §2.2CLP-1 §1.5Key results
Worked solution
Strategy
An exponential behaves differently at the two ends: eˣ grows without bound as x → ∞ but tends to 0 as x → −∞. On the left, then, the constant terms are the ones that matter.
Working
For very negative x, eˣ is a tiny positive number.
The bottom tends to 3, not 0, so the limit is the quotient of the two limits.
For the whole picture: on the right, divide by eˣ and the terms in e−x vanish, so the other asymptote is y = 4.
Answer
The answer is y = −5/3, about −1.667.
Check
At x = −10, eˣ is under 0.0001 and f ≈ −1.6665, close to −5/3 ≈ −1.6667.
Where this goes wrong
Dividing the top and the bottom by eˣ and answering 4 for both ends. That is right as x → ∞, but as x → −∞ the terms in e−x grow without bound; there eˣ → 0 and the constants give −5/3.
Leans on
The graph of the exponential function, from Functions, graphs and inverses
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