Sunday, 10 July 2016

calculus - Evaluation of limxto0fracsin(x+a)sin(a)sin2x.



I am having trouble proving out the below:




lim





I have got as far as below using \sin(a) + \sin(b) in numerator and \sin(2a) in the denominator but am not sure how to expand further, or if these are the right rules to apply.



\lim_{x \to 0}\frac{\sin(x)\cos(a) + \sin(a)[\cos(x) - 1]}{2\sin(x)\cos(x)}



I need to simplify it down to apply \displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}.


Answer



Continue with this:
\frac{\sin x \cos a}{2\sin x\cos x} + \frac{\sin a[\cos x - 1]}{2\sin x\cos x}

\frac{\cos a}{2\cos x} - \sin a\frac{2\sin^2\frac{x}{2}}{4\sin\frac{x}{2}\cos\frac{x}{2}\cos x}
\frac{\cos a}{2\cos x} - \sin a\frac{\sin\frac{x}{2}}{2\cos\frac{x}{2}\cos x}
then
\lim_{x\to0}\frac{\cos a}{2\cos x} - \sin a\frac{\sin\frac{x}{2}}{2\cos\frac{x}{2}\cos x}=\frac{\cos a}{2\times1}-0=\frac{\cos a}{2}


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...