Physics and Mathematics
Limits of the form 1^ infinity: Example 3
Practice Questions with Step-by-Step Solutions
Exponential Limits of the Form
[\lim_{x \to 0} \left(\dfrac{1 + ax^2}{1 + bx^2}\right)^{\dfrac{1}{x^2}}]
Standard Result Used (Very Important)
If
[\lim_{x \to 0} \left(\dfrac{1 + ax^2}{1 + bx^2}\right)^{\dfrac{1}{x^2}}]
Then the limit equals:
[\boxed{e^{a-b}}]
This result will be derived step-by-step in every solution below.
Question 1. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + 5x^2}{1 + 3x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Identify the form
As [x → 0],
[1 + 5x^2 → 1]
[1 + 3x^2 → 1]
So base → 1 and exponent → ∞
Form = [1^{\infty}] (indeterminate)
Step 2: Split the fraction
[\left(\dfrac{1 + 5x^2}{1 + 3x^2}\right)^{\dfrac{1}{x^2}} ][= \dfrac{(1 + 5x^2)^{\dfrac{1}{x^2}}}{(1 + 3x^2)^{\dfrac{1}{x^2}}}]
Step 3: Rewrite each term
[(1 + 5x^2)^{\dfrac{1}{x^2}} ][= \left((1 + 5x^2)^{\dfrac{1}{5x^2}}\right)^5 → e^5]
[(1 + 3x^2)^{\dfrac{1}{x^2}} ][= \left((1 + 3x^2)^{\dfrac{1}{3x^2}}\right)^3 → e^3]
Step 4: Take ratio
[\dfrac{e^5}{e^3} = e^{5-3}]
Final Answer:
[e^2]
Question 2. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + 7x^2}{1 + 2x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Form = [1^{\infty}]
Step 2: Split expression
[\dfrac{(1 + 7x^2)^{\dfrac{1}{x^2}}}{(1 + 2x^2)^{\dfrac{1}{x^2}}}]
Step 3: Apply standard limit
[(1 + 7x^2)^{\dfrac{1}{x^2}} → e^7]
[(1 + 2x^2)^{\dfrac{1}{x^2}} → e^2]
Step 4: Combine
[e^{7-2}]
Final Answer:
[e^5]
Question 3. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + 4x^2}{1 + x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Identify [a = 4], [b = 1]
Step 2: Use standard result
[e^{a-b} = e^{4-1}]
Final Answer:
[e^3]
Question 4. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + 9x^2}{1 + 5x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Examine the form
As [x → 0]:
[1 + 9x^2 → 1]
[1 + 5x^2 → 1]
[\dfrac{1}{x^2} → \infty]
So the limit is of the indeterminate form:
[\boxed{1^{\infty}}]
Step 2: Rewrite the expression by separating numerator and denominator
[\left(\dfrac{1 + 9x^2}{1 + 5x^2}\right)^{\dfrac{1}{x^2}}
= \dfrac{(1 + 9x^2)^{\dfrac{1}{x^2}}}{(1 + 5x^2)^{\dfrac{1}{x^2}}}]
Step 3: Handle the numerator
Rewrite the exponent:
[(1 + 9x^2)^{\dfrac{1}{x^2}}
= \left((1 + 9x^2)^{\dfrac{1}{9x^2}}\right)^9]
Using the standard limit:
[\lim_{u \to 0} (1 + u)^{\dfrac{1}{u}} = e]
So,
[(1 + 9x^2)^{\dfrac{1}{9x^2}} → e]
Hence,
[(1 + 9x^2)^{\dfrac{1}{x^2}} → e^9]
Step 4: Handle the denominator
Similarly,
[(1 + 5x^2)^{\dfrac{1}{x^2}}
= \left((1 + 5x^2)^{\dfrac{1}{5x^2}}\right)^5]
So,
[(1 + 5x^2)^{\dfrac{1}{5x^2}} → e]
Therefore,
[(1 + 5x^2)^{\dfrac{1}{x^2}} → e^5]
Step 5: Take the ratio
[\dfrac{e^9}{e^5} = e^{9-5}]
Final Answer:
[e^4]
Question 5. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + 3x^2}{1 + 8x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Form = [1^{\infty}]
Step 2: Separate numerator and denominator
[\dfrac{(1 + 3x^2)^{\dfrac{1}{x^2}}}{(1 + 8x^2)^{\dfrac{1}{x^2}}}]
Step 3: Rewrite numerator
[(1 + 3x^2)^{\dfrac{1}{x^2}}
= \left((1 + 3x^2)^{\dfrac{1}{3x^2}}\right)^3 → e^3]
Step 4: Rewrite denominator
[(1 + 8x^2)^{\dfrac{1}{x^2}}
= \left((1 + 8x^2)^{\dfrac{1}{8x^2}}\right)^8 → e^8]
Step 5: Take ratio
[\dfrac{e^3}{e^8} = e^{3-8}]
Final Answer:
[\dfrac{1}{e^5}]
Question 6. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + 6x^2}{1 + 6x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Both numerator and denominator are identical.
Step 2: Rewrite
[\left(\dfrac{1 + 6x^2}{1 + 6x^2}\right)^{\dfrac{1}{x^2}}
= (1)^{\dfrac{1}{x^2}}]
Step 3: Any power of 1 is 1
Final Answer:
[1]
Question 7. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + 2x^2}{1 + 5x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Form = [1^{\infty}]
Step 2: Separate terms
[\dfrac{(1 + 2x^2)^{\dfrac{1}{x^2}}}{(1 + 5x^2)^{\dfrac{1}{x^2}}}]
Step 3: Rewrite numerator
[(1 + 2x^2)^{\dfrac{1}{x^2}}
= \left((1 + 2x^2)^{\dfrac{1}{2x^2}}\right)^2 → e^2]
Step 4: Rewrite denominator
[(1 + 5x^2)^{\dfrac{1}{x^2}}
= \left((1 + 5x^2)^{\dfrac{1}{5x^2}}\right)^5 → e^5]
Step 5: Take ratio
[\dfrac{e^2}{e^5} = e^{-3}]
Final Answer:
[\dfrac{1}{e^3}]
Question 8. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + 10x^2}{1 + 4x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Form = [1^{\infty}]
Step 2: Separate expression
[\dfrac{(1 + 10x^2)^{\dfrac{1}{x^2}}}{(1 + 4x^2)^{\dfrac{1}{x^2}}}]
Step 3: Numerator
[(1 + 10x^2)^{\dfrac{1}{x^2}}
= \left((1 + 10x^2)^{\dfrac{1}{10x^2}}\right)^{10} → e^{10}]
Step 4: Denominator
[(1 + 4x^2)^{\dfrac{1}{x^2}}
= \left((1 + 4x^2)^{\dfrac{1}{4x^2}}\right)^4 → e^4]
Step 5: Take ratio
[e^{10-4}]
Final Answer:
[e^6]
Question 9. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + \pi x^2}{1 + x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: Treat [\pi] as a constant
Step 2: Separate expression
[\dfrac{(1 + \pi x^2)^{\dfrac{1}{x^2}}}{(1 + x^2)^{\dfrac{1}{x^2}}}]
Step 3: Numerator
[(1 + \pi x^2)^{\dfrac{1}{x^2}}
= \left((1 + \pi x^2)^{\dfrac{1}{\pi x^2}}\right)^{\pi} → e^{\pi}]
Step 4: Denominator
[(1 + x^2)^{\dfrac{1}{x^2}} → e]
Step 5: Take ratio
[e^{\pi – 1}]
Final Answer:
[e^{\pi – 1}]
Question 10. Evaluate [\lim_{x \to 0} \left(\dfrac{1 + (\sin 1)x^2}{1 + 2x^2}\right)^{\dfrac{1}{x^2}}]
Step-by-Step Solution:
Step 1: [\sin 1] is a constant
Step 2: Separate expression
[\dfrac{(1 + (\sin 1)x^2)^{\dfrac{1}{x^2}}}{(1 + 2x^2)^{\dfrac{1}{x^2}}}]
Step 3: Numerator
[(1 + (\sin 1)x^2)^{\dfrac{1}{x^2}}
= \left((1 + (\sin 1)x^2)^{\dfrac{1}{(\sin 1)x^2}}\right)^{\sin 1} → e^{\sin 1}]
Step 4: Denominator
[(1 + 2x^2)^{\dfrac{1}{x^2}} → e^2]
Step 5: Take ratio
[e^{\sin 1 – 2}]
Final Answer:
[e^{\sin 1 – 2}]
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