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

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

Grade:12

5 Answers

ng29
1698 Points
8 years ago
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 zero
bcoz gif of any nimber less than 1 will be 0 so value of limit is 0 
 
approve if useful
ng29
1698 Points
8 years ago
sinx/x  
 
it was missed in prev answer
ng29
1698 Points
8 years ago
sinx/x is less than 1
tanx/x is greater than 1
 
it was missed in prev answer
grenade
2061 Points
8 years ago
the value of lt(x->0)sinx/x is equeal to 1 from the left side this meant that the value might be 0.9999999999999
and [ 0.9999 ] = 0 therefore [sinx/x] =0
Azhar
13 Points
3 years ago
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

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