JEE Class main Answered
Sir please provide answer to this question
Asked by anshuman.anshuman090 | 19 Sep, 2020, 12:05: PM
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, 07:38: PM
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