Physics and Mathematics
Limits Form 4: Exponential Limits
1. Standard Exponential Limit
The most important exponential limit is:
[\lim_{x→0} \dfrac{(e^{x}-1)}{x} = 1]
Why this limit is important
- It is fundamental for:
- Limits involving [e^{ax}]
- Differentiation of exponential functions
- It is treated as a standard result (similar to trigonometric standard limits)
How to Use This Standard Limit
If the given expression can be converted into the form
[\dfrac{(e^{θ}-1)}{θ}],
then its limit as [θ→0] is 1.
Practice Questions – Exponential Limits
Question 1. Evaluate: [\lim_{x→0} \dfrac{(e^{x}-1)}{x}]
Step-by-Step Solution:
As [x→0]:
[e^{x} → 1]
Hence, [e^{x}-1 → 0]
[x → 0]
So the limit is of the form [\dfrac{0}{0}].
We use the standard exponential limit:
[\lim_{x→0} \dfrac{(e^{x}-1)}{x} = 1]
Final Answer:
[1]
Question 2. Evaluate: [\lim_{x→0} \dfrac{(e^{2x}-1)}{x}]
Step-by-Step Solution:
As [x→0], the expression gives [\dfrac{0}{0}].
Rewrite the expression:
[\dfrac{e^{2x}-1}{x} ][= 2 \cdot \dfrac{e^{2x}-1}{2x}]
As [x→0], [2x→0].
Apply the standard limit:
[\dfrac{e^{2x}-1}{2x} → 1]
Multiply:
[2 \cdot 1 = 2]
Final Answer:
[2]
Question 3. Evaluate: [\lim_{x→0} \dfrac{(e^{5x}-1)}{x}]
Step-by-Step Solution:
The form is [\dfrac{0}{0}].
Rewrite:
[\dfrac{e^{5x}-1}{x} ][= 5 \cdot \dfrac{e^{5x}-1}{5x}]
As [x→0], [5x→0].
Apply:
[\dfrac{e^{5x}-1}{5x} → 1]
Multiply:
[5 \cdot 1 = 5]
Final Answer:
[5]
Question 4. Evaluate: [\lim_{x→0} \dfrac{(e^{3x}-1)}{2x}]
Step-by-Step Solution:
Rewrite the expression:
[\dfrac{e^{3x}-1}{2x} ][= \dfrac{3}{2} \cdot \dfrac{e^{3x}-1}{3x}]
As [x→0], [3x→0].
Apply the standard limit:
[\dfrac{e^{3x}-1}{3x} → 1]
Multiply:
[\dfrac{3}{2} \cdot 1 = \dfrac{3}{2}]
Final Answer:
[\dfrac{3}{2}]
Question 5. Evaluate: [\lim_{x→0} \dfrac{(e^{4x}-1)}{5x}]
Step-by-Step Solution:
Rewrite:
[\dfrac{e^{4x}-1}{5x} ][= \dfrac{4}{5} \cdot \dfrac{e^{4x}-1}{4x}]
As [x→0], [4x→0].
Apply:
[\dfrac{e^{4x}-1}{4x} → 1]
Multiply:
[\dfrac{4}{5}]
Final Answer:
[\dfrac{4}{5}]
Question 6. Evaluate: [\lim_{x→0} \dfrac{(e^{-x}-1)}{x}]
Step-by-Step Solution:
As [x→0], the form is [\dfrac{0}{0}].
Rewrite:
[\dfrac{e^{-x}-1}{x} ][= \dfrac{e^{-x}-1}{-x}]
As [x→0], [-x→0].
Apply the standard limit:
[\dfrac{e^{-x}-1}{-x} → 1]
Hence:
[-1]
Final Answer:
[-1]
Question 7. Evaluate: [\lim_{x→0} \dfrac{(e^{6x}-1)}{3x}]
Step-by-Step Solution:
Rewrite:
[\dfrac{e^{6x}-1}{3x} ][= 2 \cdot \dfrac{e^{6x}-1}{6x}]
As [x→0], [6x→0].
Apply standard limit:
[\dfrac{e^{6x}-1}{6x} → 1]
Multiply:
[2]
Final Answer:
[2]
Question 8. Evaluate: [\lim_{x→0} \dfrac{(e^{x}-1)}{3x}]
Step-by-Step Solution:
Rewrite:
[\dfrac{1}{3} \cdot \dfrac{e^{x}-1}{x}]
Apply standard limit:
[\dfrac{e^{x}-1}{x} → 1]
Multiply:
[\dfrac{1}{3}]
Final Answer:
[\dfrac{1}{3}]
Question 9. Evaluate: [\lim_{x→0} \dfrac{(e^{7x}-1)}{14x}]
Step-by-Step Solution:
Rewrite:
[\dfrac{7}{14} \cdot \dfrac{e^{7x}-1}{7x}]
Simplify:
[\dfrac{1}{2} \cdot \dfrac{e^{7x}-1}{7x}]
Apply standard limit:
[\dfrac{1}{2}]
Final Answer:
[\dfrac{1}{2}]
Question 10. Evaluate: [\lim_{x→0} \dfrac{(e^{9x}-1)}{3x}]
Step-by-Step Solution:
Rewrite:
[3 \cdot \dfrac{e^{9x}-1}{9x}]
As [x→0], [9x→0].
Apply:
[\dfrac{e^{9x}-1}{9x} → 1]
Multiply:
[3]
Final Answer:
[3]
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