Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
201 views
in Limit, continuity and differentiability by (29 points)
recategorized by
\( \operatorname{lic}_{x \rightarrow 0} \frac{1-\cos 3 x \cos 9 x \cos 27 x-\cdots-\cos \left(3^{n} x\right)}{1-\cos \frac{x}{3} \cos \frac{x}{9} \cos \frac{x}{27} \cdots \cos \frac{x}{3^{n}}}=3^{10} \)

Please log in or register to answer this question.

1 Answer

0 votes
by (53.4k points)

For n = 1

\(\lim\limits_{x\to 0} \cfrac{1 - cos 3x}{1 - cos \frac x3} \)   \(\left(\frac 00 - case\right)\)

\(= \lim\limits_{x \to 0} \cfrac {3sin \,3x}{\frac 13 sin\frac x3}\)

\(= \lim\limits_{x\to 0} \cfrac{3^2 \frac {sin 3x}{3x}. \,3x}{\frac{sin \frac x3}{\frac x3}.\,\frac x3}\)

\(= \lim\limits_{x \to 0} 3^4 \;\frac {sin 3x}{3x}\; \frac {\frac x3}{sin\frac x3}\)

\( = 3^4 \ne 3^{10}\)

For n = 2

\(\lim\limits_{x\to0} \cfrac {1 - cos3x \,cos 9x}{1 - cos \frac x3\, cos \frac x9}\)

\(= \lim\limits_{x \to 0} \cfrac{9cos 3x \,sin 9x + 3sin3x \,cos 9x}{\frac 19 cos \frac x3\, sin\frac x9 + \frac 13 sin\frac x3\,cos\frac x9}\)

\(= \cfrac {81 + 9}{\frac 1{81} + \frac 19}\)

\(= \cfrac {90}{\frac {9+1}{81}} \)

\(= 90 \times \frac{81}{10} \)

\(= 729 \)

\(= 3^6\)

For n = 3

\(\lim\limits_{x\to 0} \cfrac {1 - cos 3x \,cos 9x\,cos 27 x}{1 - cos\frac x3 \, cos \frac x9\, cos \frac x{27}}\)

\(= \lim\limits_{x\to 0} \cfrac {3\,sin 3x \, cos 9x\, cos27x + 9\,cos3x \, sin9x\, cos27x + 27\,cos3x \, cos 9x \, sin 27x \, }{\frac 13 \,sin \frac x3 \, cos \frac x9 \, cos \frac x{27} + \frac 19\, cos \frac x3 \, sin \frac x9 \, cos \frac x{27}+ \frac 1{27}\,cos \frac x3\, cos \frac x9 \, sin \frac x{27}}\)

\(=\cfrac {9 + 81 + 729}{\frac 1{9} + \frac 1{81} + \frac 1{729}}\)

\(= \cfrac {819}{\frac{81 + 9 + 1}{729}}\)

\(= 819 \times \frac{729}{91} \)

\(= 9 \times 729 \)

\(= 3^8\)

For n = 4

\(\lim\limits_{x\to 0} \cfrac {1 - cos 3x \, cos 9x\, cos 27x\, cos 81x}{1 - cos \frac x3 \, cos \frac x9 \, cos\frac x{27} \, cos\frac x{81}}\)

\(= \cfrac{3^2 + 9^2 + 27^2 + 81^2}{\left(\frac13\right)^2 + \left(\frac19\right)^2 + \left(\frac1{27}\right)^2 + \left(\frac1{81}\right)^2}\)

\(= \cfrac {9 + 81 + 729 + 6561}{ \frac 19 + \frac 1{81} + \frac 1{729} + \frac 1{6561}}\)

\(= \cfrac {7380}{\frac{729 + 81 + 9 + 1}{6561}}\)

\(= 7380 \times \frac {6561}{820}\)

\(= 9 \times 6561\)

\(= 3^2 \times 3^8\)

\(= 3^{10}\)

Hence, the value of n is 4.

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...