 Click to Chat

1800-1023-196

+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
```        limit x tends to 0 ((x - sin  x)/x)sin(1/x)
A.1
B.does not exist
C.-1
D.0```
7 years ago

```							D.0  use expansion of sinx
```
7 years ago
```							lim x-->0 [(x-sinx)/x][sin(1/x)]
LHL->  lim h-->0 [(-h+sin h)/(-h)][sin(1/(-h)]
[((sin h)- h)/h][sin(1/h)]
[((sinh)/h)-1][sin(1/h)]
[sin(1/h)]=0                       {lim h-->0 (sin h)/h  =1}
RHL->  lim h-->0  [(h-sin h)/h][sin(1/h)]
[1- (sin h)/h][sin(1/h)]
[sin(1/h)]=0                          {"   "   "    "    "     "   "}
RHL=LHL=0
```
7 years ago
```							D.0
by using algebra of limits
=lim x tends to 0 (x/x - sinx/x) lim x tends to 0 sin(1/x).x/x   (multiply and divide by x)
=(1-1).0
=0
```
7 years ago
```							the concept is that the sin(infinity) has a finite value between -1 and 1 so cannot be treated as indeterminate form. so simplyfying and putting the limit we have 0*sin(infinity) which is 0
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Differential Calculus

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 51 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions