Limx—0 1-cos4x/x²
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Amit Pandey 8 years, 6 months ago
- 1 answers
Test
Related Questions
Posted by Xxxxxx Xx 1 year, 4 months ago
- 0 answers
Posted by Xxxxxx Xx 1 year, 4 months ago
- 3 answers
Posted by Charu Baid 1 year, 4 months ago
- 0 answers
Posted by Sanjay Kumar 1 year, 5 months ago
- 0 answers
Posted by Sanjna Gupta 1 year, 4 months ago
- 4 answers
Posted by Sneha Pandey 1 year, 5 months ago
- 0 answers
Posted by Karan Kumar Mohanta 1 year, 5 months ago
- 0 answers
Posted by Ananya Singh 1 year, 6 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Naveen Sharma 8 years, 6 months ago
Ans. {tex}\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 4x}}{{{x^2}}}{/tex}
We know {tex}1 - cos \ 2\theta = 2sin^2 \theta{/tex}
So {tex}1- cos4x = 2sin^22x{/tex}
So Now,
{tex}\mathop {\lim }\limits_{x \to 0} \frac{{2sin^22x}}{{{x^2}}}{/tex}
= {tex}2\mathop {\lim }\limits_{x \to 0} \frac{{4sin^22x}}{{{4x^2}}}{/tex}
= {tex}8\mathop {\lim }\limits_{x \to 0} \frac{{sin2x}}{{{2x}}}\times {sin2x\over 2x}{/tex}
= {tex}8\times 1 \times 1 = 8 {/tex}
0Thank You