JEE Main 2024 8 April Morning Shift
Exam JEE Main
Year 2024
Date 8 April
Shift Morning Shift
Subject Mathematics
Chapter Limits JEE Main 2024 · 8 April · Morning Shift · Mathematics · Limits
The value of _ x arrow 0 2 ( 1- x 2 x [3] 3 x [10] 10 x x^2 )is
The value of
lim x → 0 2 ( 1 − cos x cos 2 x cos 3 x 3 … … cos 10 x 10 x 2 ) i s \lim _{x \rightarrow 0}2 \left(\frac{1-\cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \ldots \ldots \sqrt[10]{\cos 10 x}}{x^2}\right)is lim x → 0 2 ( x 2 1 − c o s x c o s 2 x 3 c o s 3 x …… 10 c o s 10 x ) i s
Step-by-step solution
lim x → 0 2 ( 1 − ( 1 − x 2 2 ! ) ( 1 − 4 x 2 2 ! ) ( 1 − 9 x 2 2 ! ) … . ( 1 − 100 x 2 2 ! ) x 2 ) \lim _{x \rightarrow 0} 2\left(\frac{1-\left(1-\frac{x^2}{2 !}\right)\left(1-\frac{4 x^2}{2 !}\right)\left(1-\frac{9 x^2}{2 !}\right) \ldots .\left(1-\frac{100 x^2}{2 !}\right)}{x^2}\right) lim x → 0 2 ( x 2 1 − ( 1 − 2 ! x 2 ) ( 1 − 2 ! 4 x 2 ) ( 1 − 2 ! 9 x 2 ) … . ( 1 − 2 ! 100 x 2 ) ) By expansion
lim x → 0 2 ( 1 − ( 1 − x 2 2 ) ) ( 1 − 1 2 ⋅ 4 x 2 2 ) ( 1 − 1 3 ⋅ 9 x 2 2 ) … . . ( 1 − 1 10 ⋅ 100 x 2 2 ) x 2 ) lim x → 0 2 ( 1 − ( 1 − x 2 2 ) ( 1 − 2 x 2 2 ) ( 1 − 3 x 2 2 ) … . . ( 1 − 10 x 2 2 ) ) x 2 ) lim x → 0 2 ( 1 − 1 + x 2 ( 1 2 + 2 2 + 3 2 + … . + 10 2 ) ) x 2 2 ( 1 2 + 2 2 + 3 2 + … . + 10 2 ) 1 + 2 + … … + 10 = 10 × 11 2 = 55 \begin{gathered}
\left.\lim _{x \rightarrow 0} \frac{2\left(1-\left(1-\frac{x^2}{2}\right)\right)\left(1-\frac{1}{2} \cdot \frac{4 x^2}{2}\right)\left(1-\frac{1}{3} \cdot \frac{9 x^2}{2}\right) \ldots . .\left(1-\frac{1}{10} \cdot \frac{100 x^2}{2}\right)}{x^2}\right) \\
\lim _{x \rightarrow 0} 2\left(\frac{\left.1-\left(1-\frac{x^2}{2}\right)\left(1-\frac{2 x^2}{2}\right)\left(1-\frac{3 x^2}{2}\right) \ldots . .\left(1-\frac{10 x^2}{2}\right)\right)}{x^2}\right) \\
\lim _{x \rightarrow 0} \frac{2\left(1-1+x^2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots .+\frac{10}{2}\right)\right)}{x^2} \\
2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots .+\frac{10}{2}\right) \\
1+2+\ldots \ldots+10=\frac{10 \times 11}{2}=55
\end{gathered} x → 0 lim x 2 2 ( 1 − ( 1 − 2 x 2 ) ) ( 1 − 2 1 ⋅ 2 4 x 2 ) ( 1 − 3 1 ⋅ 2 9 x 2 ) … .. ( 1 − 10 1 ⋅ 2 100 x 2 ) x → 0 lim 2 x 2 1 − ( 1 − 2 x 2 ) ( 1 − 2 2 x 2 ) ( 1 − 2 3 x 2 ) … .. ( 1 − 2 10 x 2 ) ) x → 0 lim x 2 2 ( 1 − 1 + x 2 ( 2 1 + 2 2 + 2 3 + … . + 2 10 ) ) 2 ( 2 1 + 2 2 + 2 3 + … . + 2 10 ) 1 + 2 + …… + 10 = 2 10 × 11 = 55
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