Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 3,180 off
USE CODE: CHEM20

```Solved Examples on Limits

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) = 3and 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

```
• Complete AIPMT/AIIMS Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 3,180 off
USE CODE: CHEM20