Physics and Mathematics
Limits Form 4: Exponential Limits Example 2
Practice Questions – Exponential Limits of the Form
[\dfrac{e^{ax} – e^{bx}}{x}]
Question 1. Evaluate: [\lim_{x→0} \dfrac{e^{3x} – e^{2x}}{x}]
Step-by-Step Solution:
As [x→0]:
[e^{3x} → 1]
[e^{2x} → 1]
Hence, numerator → [1 − 1 = 0] and denominator → [0].
So the limit is of the form [\dfrac{0}{0}].
Rewrite the numerator by adding and subtracting 1:
[\dfrac{(e^{3x}-1) – (e^{2x}-1)}{x}]
Split the fraction:
[\dfrac{e^{3x}-1}{x} – \dfrac{e^{2x}-1}{x}]
Use the standard result:
[\lim_{x→0} \dfrac{e^{3x}-1}{x} = 3]
[\lim_{x→0} \dfrac{e^{2x}-1}{x} = 2]
Subtract the results:
[3 − 2 = 1]
Final Answer:
[1]
Question 2. Evaluate: [\lim_{x→0} \dfrac{e^{5x} – e^{x}}{x}]
Step-by-Step Solution:
As [x→0], both [e^{5x}] and [e^{x}] approach [1].
Hence, the form is [\dfrac{0}{0}].
Rewrite the numerator:
[\dfrac{(e^{5x}-1) – (e^{x}-1)}{x}]
Separate the terms:
[\dfrac{e^{5x}-1}{x} – \dfrac{e^{x}-1}{x}]
Apply the standard limit:
[\dfrac{e^{5x}-1}{x} → 5]
[\dfrac{e^{x}-1}{x} → 1]
Subtract:
[5 − 1 = 4]
Final Answer:
[4]
Question 3. Evaluate: [\lim_{x→0} \dfrac{e^{4x} – e^{2x}}{x}]
Step-by-Step Solution:
As [x→0]:
[e^{4x} → 1]
[e^{2x} → 1]
So the limit is of the form [\dfrac{0}{0}].
Rewrite the expression:
[\dfrac{(e^{4x}-1) – (e^{2x}-1)}{x}]
Split the fraction:
[\dfrac{e^{4x}-1}{x} – \dfrac{e^{2x}-1}{x}]
Apply the standard result separately:
[\lim_{x→0} \dfrac{e^{4x}-1}{x} = 4]
[\lim_{x→0} \dfrac{e^{2x}-1}{x} = 2]
Subtract:
[4 − 2 = 2]
Final Answer:
[2]
Question 4. Evaluate: [\lim_{x→0} \dfrac{e^{7x} – e^{3x}}{x}]
Step-by-Step Solution:
As [x→0], the expression becomes [\dfrac{0}{0}].
Rewrite:
[\dfrac{(e^{7x}-1) – (e^{3x}-1)}{x}]
Split:
[\dfrac{e^{7x}-1}{x} – \dfrac{e^{3x}-1}{x}]
Apply standard limits:
[7]
[3]
Subtract:
[7 − 3 = 4]
Final Answer:
[4]
Question 5. Evaluate: [\lim_{x→0} \dfrac{e^{2x} – e^{-x}}{x}]
Step-by-Step Solution:
As [x→0]:
[e^{2x} → 1]
[e^{-x} → 1]
So the form is [\dfrac{0}{0}].
Rewrite:
[\dfrac{(e^{2x}-1) – (e^{-x}-1)}{x}]
Split:
[\dfrac{e^{2x}-1}{x} – \dfrac{e^{-x}-1}{x}]
Apply limits:
[\lim_{x→0} \dfrac{e^{2x}-1}{x} = 2]
[\lim_{x→0} \dfrac{e^{-x}-1}{x} = -1]
Subtract:
[2 − (−1) = 3]
Final Answer:
[3]
Question 6. Evaluate: [\lim_{x→0} \dfrac{e^{x} – e^{-2x}}{x}]
Step-by-Step Solution:
Form is [\dfrac{0}{0}].
Rewrite:
[\dfrac{(e^{x}-1) – (e^{-2x}-1)}{x}]
Split:
[\dfrac{e^{x}-1}{x} – \dfrac{e^{-2x}-1}{x}]
Apply limits:
[1]
[-2]
Subtract:
[1 − (−2) = 3]
Final Answer:
[3]
Question 7. Evaluate: [\lim_{x→0} \dfrac{e^{6x} – e^{4x}}{x}]
Step-by-Step Solution:
Form is [\dfrac{0}{0}].
Rewrite:
[\dfrac{(e^{6x}-1) – (e^{4x}-1)}{x}]
Split:
[\dfrac{e^{6x}-1}{x} – \dfrac{e^{4x}-1}{x}]
Apply limits:
[6]
[4]
Subtract:
[2]
Final Answer:
[2]
Question 8. Evaluate: [\lim_{x→0} \dfrac{e^{9x} – e^{5x}}{x}]
Step-by-Step Solution:
Form is [\dfrac{0}{0}].
Rewrite:
[\dfrac{(e^{9x}-1) – (e^{5x}-1)}{x}]
Split:
[\dfrac{e^{9x}-1}{x} – \dfrac{e^{5x}-1}{x}]
Apply limits:
[9]
[5]
Subtract:
[4]
Final Answer:
[4]
Question 9. Evaluate: [\lim_{x→0} \dfrac{e^{-x} – e^{-3x}}{x}]
Step-by-Step Solution:
Form is [\dfrac{0}{0}].
Rewrite:
[\dfrac{(e^{-x}-1) – (e^{-3x}-1)}{x}]
Split:
[\dfrac{e^{-x}-1}{x} – \dfrac{e^{-3x}-1}{x}]
Apply limits:
[-1]
[-3]
Subtract:
[-1 − (−3) = 2]
Final Answer:
[2]
Question 10. Evaluate: [\lim_{x→0} \dfrac{e^{8x} – e^{2x}}{x}]
Step-by-Step Solution:
Form is [\dfrac{0}{0}].
Rewrite:
[\dfrac{(e^{8x}-1) – (e^{2x}-1)}{x}]
Split:
[\dfrac{e^{8x}-1}{x} – \dfrac{e^{2x}-1}{x}]
Apply limits:
[8]
[2]
Subtract:
[6]
Final Answer:
[6]
Key Takeaway for Students
For limits of the form
[\lim_{x→0} \dfrac{e^{ax}-e^{bx}}{x}]
Direct Result:
[\boxed{a – b}]
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