Posted by Abi on Wednesday, February 9, 2011 at 9:59pm.
Use Lim x->0 sin(x)/x =0
The proof can be done using the sandwich (squeeze) theorem.
Expand
sin(3x)
=3cos²(x) sin(x)-sin³(x)
=sin(x)(3cos²(x)-1+(1-sin²(x)))
=sin(x)(3cos²(x)-1+cos²(x))
=sin(x)(4cos²(x)-1)
As x->0, cos(x)->1, so the limit becomes
Lim x->0 3sin(x)/x
=3 Lim x->0 sin(x)/x
The latter limit can be quoted as a standard limit, or can be proven using the sandwich theorem.
The first value I tried was 3... and it didn't work either, I thought that the fact that it said "x is in degrees" affected. It also said at the end: (Give your answer accurate to at least 0.01...
Sorry, I am used to working with radians without thinking when dealing with limits and calculus. I have not properly read the instructions (would be fatal in exams!).
To change from degrees to radians, we need to multiply the number of degrees by %pi;/180, so the problem in radians is really:
Lim x->0 sin(3x*π/180)/x
=Lim x-> sin(πx/60)/x
=π/60
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