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JEE Revision Notes

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Illustration 1: Find the following limit(1984)
limn→∞ [1/(1-n2) + 2/(1-n2) + … + n/(1-n2)]
Solution:limn→∞ [1/(1-n2) + 2/(1-n2) + … + n/(1-n2)]
= limn→∞[1+2+3+…. +n]/ (1-n2)
= limn→∞n(n+1)/ 2(1-n)(1+n)
= limn→∞ n/ 2(1-n)
= -1/2

Illustration 2: limx→0 sin (π cos2x) / x2equals(2001)

1. –π
2. Π

3. π/2

4. 1
Solution: We need to compute the following expression
limx→0 sin (π cos2x) / x2
= limx→0 sin (π - πsin2x) / x2
= limx→0 sin (π sin2x) / π sin2x .π sin2x/πx2 . π
= 1.1.π
=n

Illustration 3: let f: R→R besuch that f(1) = 3 and f’(1) = 6. The find the value of limx→0 [f(1+x)/ f(1)]1/x. (2002)
Solution: Let y = [f(1+x)/ f(1)]1/x
So, log y = 1/x[log f(1+x) – log  f(1)]
So, limx→0 log y = limx→0[1/f(1+x) . f’(1+x)]
= f’(1)/ f(1)
= 6/3
log (limx→0 y) =2
limx→0 y = e2

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