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```
limit[sinx/x] as x tends to 0 where[.] is gif

```
5 years ago

ng29
1698 Points
```							first of all u must know that  sinx/x  now according to ur question  limit[sinx/x] as x tends to 0  will be equal to zerobcoz gif of any nimber less than 1 will be 0 so value of limit is 0  approve if useful
```
5 years ago
ng29
1698 Points
```							sinx/x   it was missed in prev answer
```
5 years ago
ng29
1698 Points
```							sinx/x is less than 1tanx/x is greater than 1 it was missed in prev answer
```
5 years ago
2061 Points
```							the value of lt(x->0)sinx/x is equeal to 1 from the left side this meant that the value might be 0.9999999999999and [ 0.9999 ] = 0 therefore [sinx/x] =0
```
5 years ago
Azhar
13 Points
```							Expand sinx as  [(x - x^3/3 +x^5/5 ----------)/x]Now                =>[(1 - x^2/3 +x^4/5 ----------)] Put lim x->0  =>[1 -0 -0 -----]So ans = 1
```
8 months ago
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### Course Features

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