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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
lim⁡x→0729x−243x−81x+9x+3x−1x3=K(log⁡3)3\displaystyle\lim_{x\to 0}\dfrac{729^{x}-243^{x}-81^{x}+9^{x}+3^{x}-1}{x^{3}}=K(\log 3)^{3}. Then K= ?K=\,?
A44
B55
C66correct
D77
Solution
Step 1: Express each term as a power of 33:
729x=36x,  243x=35x,  81x=34x,  9x=32x,  3x=3x729^{x}=3^{6x},\;243^{x}=3^{5x},\;81^{x}=3^{4x},\;9^{x}=3^{2x},\;3^{x}=3^{x}
 Step 2: Group cleverly to factor the numerator:
=35x(3x−1)−32x(32x−1)+(3x−1)=3^{5x}(3^{x}-1)-3^{2x}(3^{2x}-1)+(3^{x}-1)
 Step 3: Use 32x−1=(3x−1)(3x+1)3^{2x}-1=(3^{x}-1)(3^{x}+1) and pull out (3x−1)(3^{x}-1):
=(3x−1)[35x−33x−32x+1]=(3^{x}-1)\left[3^{5x}-3^{3x}-3^{2x}+1\right]
 Step 4: Factor the bracket: 35x−33x−32x+1=(32x−1)(33x−1)3^{5x}-3^{3x}-3^{2x}+1=(3^{2x}-1)(3^{3x}-1). Step 5: So the numerator factors as (3x−1)(9x−1)(27x−1)(3^{x}-1)(9^{x}-1)(27^{x}-1). Step 6: Split the limit:
lim⁡x→03x−1x⋅lim⁡x→09x−1x⋅lim⁡x→027x−1x\lim_{x\to 0}\dfrac{3^{x}-1}{x}\cdot\lim_{x\to 0}\dfrac{9^{x}-1}{x}\cdot\lim_{x\to 0}\dfrac{27^{x}-1}{x}
 Step 7: Use lim⁡x→0ax−1x=log⁡a\displaystyle\lim_{x\to 0}\dfrac{a^{x}-1}{x}=\log a:
=log⁡3⋅log⁡9⋅log⁡27=log⁡3⋅2log⁡3⋅3log⁡3=6(log⁡3)3=\log 3\cdot\log 9\cdot\log 27=\log 3\cdot 2\log 3\cdot 3\log 3=6(\log 3)^{3}
 Step 8: Compare with K(log⁡3)3K(\log 3)^{3}, so K=6K=6. Correct answer: (3) 66
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