Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

Sir please provide answer to this question 
question image
Asked by anshuman.anshuman090 | 19 Sep, 2020, 12:05: PM
answered-by-expert Expert Answer
TO find the limit x tends to -1 for the function f(x) = {(1+x)(1-x2)(1+x3)(1-x4)...(1-x4n)}/{[(1+x)(1-x2)(1+x3)(1-x4)...(1-x2n)]2}
Cancel out the common terms in the numerator and denominator so we get 
f(x) = {(1+x2n+1)(1-x2n+2)(1+x2n+3)(1-x2n+4)...(1-x4n)}/{(1+x)(1-x2)(1+x3)(1-x4)...(1-x2n)}
Now use (a2 - b2) = (a - b)(a + b)
Again cancel out the common terms
Now substitute x = -1
If the value coming out to be 0, then again use the identity (a2 - b2) = (a + b)(a - b) and cancel out common terms
COntinue this till we get non-zero value
Answered by Renu Varma | 21 Sep, 2020, 19:38: PM
JEE main - Maths
Asked by pallavibirangal87 | 08 Dec, 2024, 10:18: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
Asked by pritisindra6 | 06 Dec, 2024, 21:04: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
Asked by pawandhayal125 | 06 Oct, 2024, 10:03: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
Asked by srishtipra2008 | 28 Sep, 2024, 10:33: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
Asked by yogeshsoni575 | 23 Sep, 2024, 23:01: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
Asked by pritamlovevanshi5 | 15 Sep, 2024, 08:24: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
Asked by ganeshbabugottapu7 | 12 Sep, 2024, 18:19: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Maths
Asked by ganeshbabugottapu7 | 10 Sep, 2024, 07:32: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT