Physics and Mathematics
L Hospital’s Rule
1. Concept Overview: What Is L’Hospital’s Rule?
L’Hospital’s Rule is a method used to evaluate limits of the form:
- [\dfrac{0}{0}]
- [\dfrac{\infty}{\infty}]
by converting them into a simpler limit involving derivatives.
Very Important:
L’Hospital’s Rule is not a universal method.
It can be applied only under specific conditions.
2. Why Do We Need L’Hospital’s Rule?
Consider a limit like:
[\lim_{x \to 0} \dfrac{\sin x}{x}]
- Both numerator and denominator → 0
- Form = [\dfrac{0}{0}]
We already know its value using standard limits.
But what about:
[\lim_{x \to 0} \dfrac{\ln(1+x)}{x}]
or
[\lim_{x \to 0} \dfrac{e^x – 1}{x}]
Such limits become difficult to evaluate directly.
This is where L’Hospital’s Rule becomes useful.
3. Statement of L’Hospital’s Rule (Exam Ready)
L’Hospital’s Rule
If:
- [\lim_{x \to a} f(x) = 0] and [\lim_{x \to a} g(x) = 0]
or - [\lim_{x \to a} f(x) = \pm \infty] and [\lim_{x \to a} g(x) = \pm \infty]
and if:
- [f(x)] and [g(x)] are differentiable near [x = a]
- [g'(x) \neq 0] near [x = a]
then:
[\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}]
provided the limit on the right-hand side exists.
4. Forms Where L’Hospital’s Rule Is Applicable
Allowed Forms
- [\dfrac{0}{0}]
- [\dfrac{\infty}{\infty}]
Not Directly Allowed Forms
- [0 \cdot \infty]
- [\infty – \infty]
- [0^0], [1^\infty], [\infty^0]
These must first be converted into either:
- [\dfrac{0}{0}] or
- [\dfrac{\infty}{\infty}]
Only then can L’Hospital’s Rule be applied.
5. Step-by-Step Procedure (Never Skip This in Exams)
Whenever you want to apply L’Hospital’s Rule, always follow this exact sequence:
Step 1: Substitute the value of [x]
Check whether the form is [\dfrac{0}{0}] or [\dfrac{\infty}{\infty}]
Step 2: Confirm differentiability
Both numerator and denominator must be differentiable.
Step 3: Differentiate numerator and denominator separately
[\dfrac{f(x)}{g(x)} ;\longrightarrow; \dfrac{f'(x)}{g'(x)}]
Do NOT apply quotient rule
Step 4: Re-evaluate the limit
- If still indeterminate, apply L’Hospital’s Rule again
- Otherwise, stop
6. Simple Example (Fully Explained)
Example 1
Evaluate:
[\lim_{x \to 0} \dfrac{e^x – 1}{x}]
Step 1: Check the form
As [x → 0]:
- Numerator → [e^0 − 1 = 0]
- Denominator → [0]
Form = [\dfrac{0}{0}]
Step 2: Apply L’Hospital’s Rule
Differentiate numerator and denominator:
- Numerator derivative: [\dfrac{d}{dx}(e^x – 1) = e^x]
- Denominator derivative: [\dfrac{d}{dx}(x) = 1]
Step 3: Evaluate the new limit
[\lim_{x \to 0} \dfrac{e^x}{1} = e^0 = 1]
Final Answer
[\boxed{1}]
7. Very Important Warnings
- Do not apply L’Hospital’s Rule blindly
- Do not use it for every [\dfrac{0}{0}] form
- Do not differentiate numerator and denominator together
- Always check the form first
- Use standard limits when possible (they are faster)
8. When NOT to Use L’Hospital’s Rule (Exam Gold)
For limits like:
- [\lim_{x \to 0} \dfrac{\sin x}{x}]
- [\lim_{x \to 0} \dfrac{\ln(1+x)}{x}]
Standard limits are preferred
Using L’Hospital’s Rule here may result in loss of conceptual clarity.
9. Practice Questions with Step-by-Step Solutions
Question 1. Evaluate [\lim_{x \to 0} \dfrac{e^x – 1}{x}]
Step-by-Step Solution:
Step 1: Check the form
As [x → 0]:
Numerator → [e^0 − 1 = 0]
Denominator → [0]
So the form is [\dfrac{0}{0}]
Step 2: Apply L’Hospital’s Rule
Differentiate numerator and denominator separately:
Numerator derivative: [\dfrac{d}{dx}(e^x – 1) = e^x]
Denominator derivative: [\dfrac{d}{dx}(x) = 1]
Step 3: Evaluate the new limit
[\lim_{x \to 0} \dfrac{e^x}{1} = e^0 = 1]
Final Answer:
[\boxed{1}]
Question 2. Find [\lim_{x \to 0} \dfrac{\ln(1+x)}{x}]
Step 1: Check the form
As [x → 0]:
[\ln(1+x) → 0]
[x → 0]
Form = [\dfrac{0}{0}]
Step 2: Apply L’Hospital’s Rule
Numerator derivative: [\dfrac{d}{dx}\ln(1+x) = \dfrac{1}{1+x}]
Denominator derivative: [1]
Step 3: Evaluate the limit
[\lim_{x \to 0} \dfrac{1}{1+x} = \dfrac{1}{1} = 1]
Final Answer:
[\boxed{1}]
Question 3. Evaluate [\lim_{x \to 0} \dfrac{\sin x}{x}] using L’Hospital’s Rule
Step 1: Check the form
As [x → 0]:
[\sin x → 0]
[x → 0]
Form = [\dfrac{0}{0}]
Step 2: Apply L’Hospital’s Rule
Numerator derivative: [\dfrac{d}{dx}(\sin x) = \cos x]
Denominator derivative: [1]
Step 3: Evaluate the limit
[\lim_{x \to 0} \cos x = \cos 0 = 1]
Final Answer:
[\boxed{1}]
Question 4. Find [\lim_{x \to 0} \dfrac{\tan x}{x}] using L’Hospital’s Rule
Step 1: Check the form
As [x → 0]:
[\tan x → 0]
[x → 0]
Form = [\dfrac{0}{0}]
Step 2: Apply L’Hospital’s Rule
Numerator derivative: [\dfrac{d}{dx}(\tan x) = \sec^2 x]
Denominator derivative: [1]
Step 3: Evaluate the limit
[\lim_{x \to 0} \sec^2 x = \sec^2 0 = 1]
Final Answer:
[\boxed{1}]
Question 5. Evaluate [\lim_{x \to 0} \dfrac{e^{2x} – 1}{x}]
Step 1: Check the form
As [x → 0]:
Numerator → [e^0 − 1 = 0]
Denominator → [0]
Form = [\dfrac{0}{0}]
Step 2: Apply L’Hospital’s Rule
Numerator derivative: [\dfrac{d}{dx}(e^{2x} – 1) = 2e^{2x}]
Denominator derivative: [1]
Step 3: Evaluate the limit
[\lim_{x \to 0} 2e^{2x} = 2e^0 = 2]
Final Answer:
[\boxed{2}]
Question 6. Find [\lim_{x \to 0} \dfrac{1 – \cos x}{x^2}]
Step 1: Check the form
As [x → 0]:
[1 − \cos x → 0]
[x^2 → 0]
Form = [\dfrac{0}{0}]
Step 2: Apply L’Hospital’s Rule (First time)
Numerator derivative: [\sin x]
Denominator derivative: [2x]
New limit:
[\lim_{x \to 0} \dfrac{\sin x}{2x}]
Step 3: Still indeterminate → Apply L’Hospital’s Rule again
Numerator derivative: [\cos x]
Denominator derivative: [2]
Step 4: Evaluate
[\lim_{x \to 0} \dfrac{\cos x}{2} = \dfrac{1}{2}]
Final Answer:
[\boxed{\dfrac{1}{2}}]
Question 7. Evaluate [\lim_{x \to 0} \dfrac{x – \sin x}{x^3}]
Step 1: Check the form
As [x → 0], numerator → 0, denominator → 0
Form = [\dfrac{0}{0}]
Step 2: First application of L’Hospital’s Rule
Numerator derivative: [1 − \cos x]
Denominator derivative: [3x^2]
New limit:
[\lim_{x \to 0} \dfrac{1 – \cos x}{3x^2}]
Step 3: Still [\dfrac{0}{0}] → Apply again
Numerator derivative: [\sin x]
Denominator derivative: [6x]
Step 4: Still [\dfrac{0}{0}] → Apply again
Numerator derivative: [\cos x]
Denominator derivative: [6]
Step 5: Evaluate
[\lim_{x \to 0} \dfrac{\cos x}{6} = \dfrac{1}{6}]
Final Answer:
[\boxed{\dfrac{1}{6}}]
Question 8. Find [\lim_{x \to \infty} \dfrac{\ln x}{x}]
Step 1: Check the form
As [x → ∞]:
[\ln x → \infty]
[x → \infty]
Form = [\dfrac{\infty}{\infty}]
Step 2: Apply L’Hospital’s Rule
Numerator derivative: [\dfrac{1}{x}]
Denominator derivative: [1]
Step 3: Evaluate
[\lim_{x \to \infty} \dfrac{1}{x} = 0]
Final Answer:
[\boxed{0}]
Question 9. Evaluate [\lim_{x \to 0} \dfrac{e^x – e^{-x}}{x}]
Step 1: Check the form
As [x → 0]:
Numerator → [1 − 1 = 0]
Denominator → [0]
Form = [\dfrac{0}{0}]
Step 2: Apply L’Hospital’s Rule
Numerator derivative: [e^x + e^{-x}]
Denominator derivative: [1]
Step 3: Evaluate
[\lim_{x \to 0} (e^x + e^{-x}) = 1 + 1 = 2]
Final Answer:
[\boxed{2}]
Question 10. Find [\lim_{x \to 0} \dfrac{\ln(1+2x) – 2x}{x^2}]
Step 1: Check the form
As [x → 0]:
Numerator → [0]
Denominator → [0]
Form = [\dfrac{0}{0}]
Step 2: First application of L’Hospital’s Rule
Numerator derivative: [\dfrac{2}{1+2x} – 2]
Denominator derivative: [2x]
Step 3: Simplify numerator
[\dfrac{2 – 2(1+2x)}{1+2x} = \dfrac{-4x}{1+2x}]
Step 4: New limit
[\lim_{x \to 0} \dfrac{-4x}{(1+2x)(2x)}]
Step 5: Cancel [x]
[\lim_{x \to 0} \dfrac{-4}{2(1+2x)}]
Step 6: Evaluate
[\dfrac{-4}{2} = -2]
Final Answer:
[\boxed{-2}]
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