Limits at infinity - exponential functions - Worksheet - Free Printable

Let’s solve each problem step by step. We’re finding limits as x → -∞ and x → ∞ for exponential functions. Remember:

- As x → ∞, e^(large positive) → ∞, e^(large negative) → 0
- As x → -∞, e^(large positive) → 0, e^(large negative) → ∞ (because the exponent becomes very negative)

We’ll go one by one.

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Problem 1: f(x) = e^(5 + 3x - x³)

Look at the exponent: 5 + 3x - x³

As x → ∞:
-x³ dominates → goes to -∞ → so exponent → -∞ → e^(-∞) → 0

As x → -∞:
-x³ → -(-∞)³ = -(-∞) = +∞ → exponent → +∞ → e^(+∞) → ∞

So:
(a) lim x→-∞ f(x) = ∞
(b) lim x→∞ f(x) = 0

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Problem 2: f(x) = e^((5x² + x)/(6 + 2x))

First, simplify the exponent: (5x² + x)/(6 + 2x)

Divide numerator and denominator by x (for large |x|):

= [5x + 1] / [6/x + 2]

As x → ∞:
Numerator → 5x → ∞
Denominator → 2
So exponent → ∞ → e^∞ → ∞

As x → -∞:
Numerator → 5x → -∞
Denominator → 2
So exponent → -∞ → e^(-∞) → 0

Wait — let’s check more carefully.

Actually, divide numerator and denominator by x:

(5x² + x)/(6 + 2x) = [x(5x + 1)] / [2x(3/x + 1)] → wait, better to factor or use leading terms.

Leading term in numerator: 5x²
Leading term in denominator: 2x
So ratio ~ (5x²)/(2x) = (5/2)x

So as x → ∞, exponent ~ (5/2)x → ∞ → e^∞ → ∞
As x → -∞, exponent ~ (5/2)x → -∞ → e^(-∞) → 0

Yes.

So:
(a) lim x→-∞ f(x) = 0
(b) lim x→∞ f(x) = ∞

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Problem 3: f(x) = 4e^(8x) - e^(-5x) - 8e^(4x)

As x → ∞:
e^(8x) grows fastest → 4e^(8x) → ∞
Other terms: e^(-5x) → 0, e^(4x) → ∞ but slower than e^(8x)
So overall → ∞

As x → -∞:
e^(8x) → 0, e^(4x) → 0
But e^(-5x) = e^(positive large) → ∞ → so -e^(-5x) → -∞
So overall → -∞

Check:
At x → -∞:
4e^(8x) ≈ 0
- e^(-5x) = -e^(|5x|) → -∞
-8e^(4x) ≈ 0
So yes, dominated by -e^(-5x) → -∞

So:
(a) lim x→-∞ f(x) = -∞
(b) lim x→∞ f(x) = ∞

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Problem 4: f(x) = 5e^(-x) - 9e^(-7x) - e^(12x)

As x → ∞:
e^(-x) → 0, e^(-7x) → 0, e^(12x) → ∞ → so -e^(12x) → -∞
Overall → -∞

As x → -∞:
e^(-x) = e^(positive) → ∞
e^(-7x) = e^(positive larger) → ∞ faster? Let’s see exponents:

When x → -∞, let x = -t, t → ∞

Then:
5e^(-x) = 5e^t
-9e^(-7x) = -9e^(7t)
-e^(12x) = -e^(-12t) → 0

Now compare 5e^t vs -9e^(7t). Since 7t > t, e^(7t) grows much faster → so -9e^(7t) dominates → → -∞

Wait — that can’t be right? Let me double-check.

Original: f(x) = 5e^(-x) - 9e^(-7x) - e^(12x)

As x → -∞:
Set x = -t, t → ∞

f(-t) = 5e^(t) - 9e^(7t) - e^(-12t)

Now:
5e^t → ∞
-9e^(7t) → -∞ (and much faster than 5e^t because 7t > t)
-e^(-12t) → 0

So dominant term is -9e^(7t) → -∞

Thus, as x → -∞, f(x) → -∞

As x → ∞:
5e^(-x) → 0
-9e^(-7x) → 0
-e^(12x) → -∞
So overall → -∞

Wait — both sides go to -∞? That seems odd but mathematically correct.

Confirm:

At x → ∞: only -e^(12x) matters → -∞
At x → -∞: -9e^(-7x) = -9e^(7|x|) → -∞, and it grows faster than 5e^(-x) = 5e^(|x|), since 7|x| > |x|

Yes.

So:
(a) lim x→-∞ f(x) = -∞
(b) lim x→∞ f(x) = -∞

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Problem 5: f(x) = (5e^(-5x) - 3e^(7x)) / (6e^(7x) - 4e^(-5x))

This is a rational function of exponentials. Best to factor out the dominant exponential in numerator and denominator depending on direction.

Case 1: x → ∞

In numerator: e^(7x) dominates over e^(-5x) → so numerator ~ -3e^(7x)
In denominator: e^(7x) dominates → denominator ~ 6e^(7x)
So f(x) ~ (-3e^(7x)) / (6e^(7x)) = -3/6 = -1/2

More precisely: divide numerator and denominator by e^(7x):

Numerator: 5e^(-5x)/e^(7x) - 3 = 5e^(-12x) - 3 → as x→∞, e^(-12x)→0 → so → -3
Denominator: 6 - 4e^(-5x)/e^(7x) = 6 - 4e^(-12x) → → 6
So limit = -3/6 = -1/2

Case 2: x → -∞

Now e^(-5x) dominates because when x is large negative, -5x is large positive.

So in numerator: 5e^(-5x) dominates over -3e^(7x) (since e^(7x)→0)
In denominator: -4e^(-5x) dominates over 6e^(7x) (which →0)

So f(x) ~ (5e^(-5x)) / (-4e^(-5x)) = 5/-4 = -5/4

More precisely: divide numerator and denominator by e^(-5x):

Numerator: 5 - 3e^(7x)/e^(-5x) = 5 - 3e^(12x) → as x→-∞, e^(12x)→0 → so → 5
Denominator: 6e^(7x)/e^(-5x) - 4 = 6e^(12x) - 4 → → -4
So limit = 5 / (-4) = -5/4

Perfect.

So:
(a) lim x→-∞ f(x) = -5/4
(b) lim x→∞ f(x) = -1/2

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Final Answers:

1. (a) ∞, (b) 0
2. (a) 0, (b) ∞
3. (a) -∞, (b) ∞
4. (a) -∞, (b) -∞
5. (a) -5/4, (b) -1/2

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Final Answer:
1. (a) ∞, (b) 0
2. (a) 0, (b) ∞
3. (a) -∞, (b) ∞
4. (a) -∞, (b) -∞
5. (a) -5/4, (b) -1/2