Courses

# Notes | EduRev

## JEE Revision Notes

Created by: Learners Habitat

## JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course JEE Revision Notes.
All you need of JEE at this link: JEE

Illustration 1: Find the following limit(1984)
limnâ†’âˆž [1/(1-n2) + 2/(1-n2) + â€¦ + n/(1-n2)]
Solution:limnâ†’âˆž [1/(1-n2) + 2/(1-n2) + â€¦ + n/(1-n2)]
= limnâ†’âˆž[1+2+3+â€¦. +n]/ (1-n2)
= limnâ†’âˆžn(n+1)/ 2(1-n)(1+n)
= limnâ†’âˆž n/ 2(1-n)
= -1/2

Illustration 2: limxâ†’0 sin (Ï€ cos2x) / x2equals(2001)

1. â€“Ï€
2. Î

3. Ï€/2

4. 1
Solution: We need to compute the following expression
limxâ†’0 sin (Ï€ cos2x) / x2
= limxâ†’0 sin (Ï€ - Ï€sin2x) / x2
= limxâ†’0 sin (Ï€ sin2x) / Ï€ sin2x .Ï€ sin2x/Ï€x2 . Ï€
= 1.1.Ï€
=n

Illustration 3: let f: Râ†’R besuch that f(1) = 3 and fâ€™(1) = 6. The find the value of limxâ†’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) â€“ log  f(1)]
So, limxâ†’0 log y = limxâ†’0[1/f(1+x) . fâ€™(1+x)]
= fâ€™(1)/ f(1)
= 6/3
log (limxâ†’0 y) =2
limxâ†’0 y = e2

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;