Calc 2 prob. Taylor Polynomial problem.?
Use the taylor polynomial of degree 5 about x=0 and sin x to explain why lim as x-->0 sin(x) / x = 1
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Use the taylor polynomial of degree 5 about x=0 and sin x to explain why lim as x-->0 sin(x) / x = 1
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f(x) = sin x ==> f(0) = 0
f '(x) = cos x ==> f '(0) = 1
f ''(x) = -sin x ==> f ''(0) = 0
f '''(x) = -cos x ==> f '''(0) = -1
f '''(x) = sin x ==> f '''(0) = 0
f '''''(x) = cos x ==> f '''''(0) = 1.
So, sin x = x - x^3/3! + x^5/5! - ...
Next,
lim(x-->0) sin(x)/x
= lim(x-->0) (1/x) * (x - x^3/3! + x^5/5! - ...)
= lim(x-->0) (1 - x^2/3! + x^4/5! - ...)
= 1 - 0 + 0 - ...
= 1.
I hope this helps!
You forgot to complete the differential using the chain rule. Do the first derivative: f '(x) = -1/(1-x)^2 * d(1-x)/dx = -1/(1-x)^2 * (-1) = + 1/(1-x)^2 Then f '(0) = 1 So the series doesn't alternate.