Sunday

December 21, 2014

December 21, 2014

Posted by **assignment** on Monday, March 5, 2012 at 1:08am.

- calculus -
**drwls**, Monday, March 5, 2012 at 1:24amI assume that the numerator is (x - sinx)

In a Taylor series,

x - sinx = x^3/3! - x^5/5! + ...

Divide that by x^3

The limit as x->0 is therefore 1/3! = 1/6

**Answer this Question**

**Related Questions**

Calculus - Work out the problem and explain your steps as you go please! 1. ...

Calculus - Evaluate the following limit: lim x->0 xcscx =limx->0 x/sinx ...

Trig Help - Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx...

Calculus - 1. Does the limit limx-->0:6[[x]] exist? 2. Does the limit limx--&...

Trig........ - I need to prove that the following is true. Thanks (cosx / 1-sinx...

Math - Prove the theorem that if limx¡úd h(x)= P and limx¡úd k(x) = Q and h(x...

calculus - Could someone check my reasoning? thanx Find the derivative of the ...

math - the problem is 2cos^2x + sinx-1=0 the directions are to "use an identity ...

Calculus - Evaluate the following integral integral 1 = a and b = 4 of sinx dx/(...

calculus - limx→∞3x+127x+11