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
Vinay Arya Grade: 12
`         Q.1. Find Limitx---->0 8/x8[1-cosx2/2 -cosx2/4 +cosx2/2 .cosx2/4]`
8 years ago

## Answers : (1)

16 Points
```
Limitx---->08/x8[(1-cosx2/2)(1-cosx2/4)]=Limitx--->08[(1-cosx2/2)/x4][(1-cosx2/4)/x4]
=Limitx---->08[2(sinx2/4)2/x4][2(sinx2/8)2/x4]=Limitx---->032[(sinx2/4)/4(x2/4)]2[(sinx2/8)/8(x2/8)]2
=Limitx--->032/(4282) [(sinx2/4)/(x2/4)]2[(sinx2/8)/(x2/8)]2  =1/32*Limitx---->0[(sinx2/4)/(x2/4)]2[(sinx2/8)/(x2/8)]2
=1/32
Here i used following formulas:
1-cosx=2sin2(x/2)
Limity--->0siny/y = 1 [replace y by x2/4 and x2/8 to get a new result which is used in solving above problem...]

```
8 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Differential Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details