Physics and Mathematics
Limits of the form 1^ infinity: Example 2
Practice Questions with Step-by-Step Solutions
Question 1. Evaluate [\lim_{x \to 0} (1 + \sin x)^{\dfrac{1}{x}}]
Step-by-Step Solution:
Step 1: Identify limits of inner functions
As [x → 0],
[\sin x → 0]
So the expression becomes of the form:
[(1 + 0)^{\infty}] = [1^{\infty}] (indeterminate form)
Step 2: Rewrite the exponent
Multiply and divide the exponent by [\sin x]:
[(1 + \sin x)^{\dfrac{1}{\sin x} \cdot \dfrac{\sin x}{x}}]
Step 3: Apply standard limits
We know:
[\lim_{u \to 0} (1 + u)^{\dfrac{1}{u}} = e]
[\lim_{x \to 0} \dfrac{\sin x}{x} = 1]
Step 4: Combine results
[e^{1}]
Final Answer:
[e]
Question 2. Evaluate [\lim_{x \to 0} (1 + \tan x)^{\dfrac{1}{x}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[\tan x → 0]
So the form is [1^{\infty}]
Step 2: Rewrite exponent carefully:
[(1 + \tan x)^{\dfrac{1}{\tan x} \cdot \dfrac{\tan x}{x}}]
Step 3: Use standard limits:
[\lim_{u \to 0} (1 + u)^{\dfrac{1}{u}} = e]
[\lim_{x \to 0} \dfrac{\tan x}{x} = 1]
Step 4: Combine:
[e^{1}]
Final Answer:
[e]
Question 3. Evaluate [\lim_{x \to 0} (1 + 2\sin x)^{\dfrac{1}{x}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[2\sin x → 0]
Form becomes [1^{\infty}]
Step 2: Rewrite expression:
[(1 + 2\sin x)^{\dfrac{1}{2\sin x} \cdot \dfrac{2\sin x}{x}}]
Step 3: Apply limits:
[(1 + u)^{\dfrac{1}{u}} → e]
[\lim_{x \to 0} \dfrac{\sin x}{x} = 1]
So exponent becomes [2]
Step 4: Final value:
[e^{2}]
Final Answer:
[e^{2}]
Question 4. Evaluate [\lim_{x \to 0} (1 + \sin 3x)^{\dfrac{1}{x}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[\sin 3x → 0]
Step 2: Rewrite:
[(1 + \sin 3x)^{\dfrac{1}{\sin 3x} \cdot \dfrac{\sin 3x}{x}}]
Step 3: Use limits:
[(1 + u)^{\dfrac{1}{u}} → e]
[\lim_{x \to 0} \dfrac{\sin 3x}{x} = 3]
Step 4: Combine:
[e^{3}]
Final Answer:
[e^{3}]
Question 5. Evaluate [\lim_{x \to 0} (1 + \tan 2x)^{\dfrac{1}{x}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[\tan 2x → 0]
Step 2: Rewrite:
[(1 + \tan 2x)^{\dfrac{1}{\tan 2x} \cdot \dfrac{\tan 2x}{x}}]
Step 3: Apply limits:
[(1 + u)^{\dfrac{1}{u}} → e]
[\lim_{x \to 0} \dfrac{\tan 2x}{x} = 2]
Step 4: Final value:
[e^{2}]
Final Answer:
[e^{2}]
Question 6. Evaluate [\lim_{x \to 0} (1 – \sin x)^{\dfrac{1}{x}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[\sin x → 0]
Step 2: Rewrite carefully:
[(1 – \sin x)^{\dfrac{1}{-\sin x} \cdot \dfrac{-\sin x}{x}}]
Step 3: Apply limits:
[(1 + u)^{\dfrac{1}{u}} → e]
[\lim_{x \to 0} \dfrac{-\sin x}{x} = -1]
Step 4: Final result:
[e^{-1}]
Final Answer:
[\dfrac{1}{e}]
Question 7. Evaluate [\lim_{x \to 0} (1 + \sin^2 x)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[\sin^2 x → 0]
Step 2: Rewrite:
[(1 + \sin^2 x)^{\dfrac{1}{\sin^2 x} \cdot \dfrac{\sin^2 x}{x^2}}]
Step 3: Use limits:
[(1 + u)^{\dfrac{1}{u}} → e]
[\lim_{x \to 0} \dfrac{\sin^2 x}{x^2} = 1]
Step 4: Final value:
[e]
Final Answer:
[e]
Question 8. Evaluate [\lim_{x \to 0} (1 + \tan^2 x)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[\tan^2 x → 0]
Step 2: Rewrite:
[(1 + \tan^2 x)^{\dfrac{1}{\tan^2 x} \cdot \dfrac{\tan^2 x}{x^2}}]
Step 3: Use:
[\lim_{x \to 0} \dfrac{\tan^2 x}{x^2} = 1]
Standard exponential limit
Step 4: Final value:
[e]
Final Answer:
[e]
Question 9. Evaluate [\lim_{x \to 0} (1 + 3\sin 2x)^{\dfrac{1}{x}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[\sin 2x → 0]
Step 2: Rewrite:
[(1 + 3\sin 2x)^{\dfrac{1}{3\sin 2x} \cdot \dfrac{3\sin 2x}{x}}]
Step 3: Apply limits:
[(1 + u)^{\dfrac{1}{u}} → e]
[\lim_{x \to 0} \dfrac{\sin 2x}{x} = 2]
Step 4: Final value:
[e^{6}]
Final Answer:
[e^{6}]
Question 10. Evaluate [\lim_{x \to 0} (1 – 4\tan x)^{\dfrac{1}{x}}]
Step-by-Step Solution:
Step 1: As [x → 0],
[\tan x → 0]
Step 2: Rewrite:
[(1 – 4\tan x)^{\dfrac{1}{-4\tan x} \cdot \dfrac{-4\tan x}{x}}]
Step 3: Use:
[(1 + u)^{\dfrac{1}{u}} → e]
[\lim_{x \to 0} \dfrac{\tan x}{x} = 1]
Step 4: Final value:
[e^{-4}]
Final Answer:
[\dfrac{1}{e^{4}}]
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