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```        lim x-->0 [cos(sinx)-cosx] / x4

ans: 1/6```
5 years ago

Swapnil Saxena
102 Points
```										Using L Hopitals rule
lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d/dx[cos(sin(x))-cos(x)]/x^4 if ([cos(sin(x))-cos(x)]=0 and x=4)
Again since the expression is yielding 0/0 appyling L hopitals rule,
lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d2/dx2[cos(sin(x))-cos(x)]/x^4 at x=0
Again since the expression is yielding 0/0 appyling L hopitals rule,
lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d3/dx3[cos(sin(x))-cos(x)]/x^4 at x=0
Again since the expression is yielding 0/0 appyling L hopitals rule,
lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d4/dx4[cos(sin(x))-cos(x)]/x^4 at x=0
Now Since the expression 4/24= 1/6 .

```
5 years ago
Chetan Mandayam Nayakar
312 Points
```										lim[cos(sinx)-cosx]/x^4
=lim[-sin(sinx)cosx+sinx]/(4x^3)
=lim[sin(sinx)sinx-cos(sinx)cos2x+cosx]/12x2
=lim[cosx-cos(sinx)cos2x]/12x2
=lim[1-cos(sinx)cosx]/12x2
=lim[sin(sinx)cos2x+cos(sinx)sinx]/24x
differentiate once more w.r.t. to get the answer
```
5 years ago
jyoti dogra
29 Points
```										thanks a lot
```
5 years ago
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