Question

What is the limit of lim_(x->0) (cos(x)-cos(5x))/x^2

Answer 1

id probably say use L hopital rule or taylor series of cos...which one is familiar for u?

Answer 2

if u are not comfortable with 2 methods i mentioned above, maybe we can use another method like substitution.

Answer 3

what do u think eugesd? :)

Answer 4

I'm used L hopitals rule, but I really want to get comfortable with Taylor Series, I'm studying for my final so I need to master everything! Could you give me an overview on how I can do this, I'm not understanding it very well in class, thanks! :)

Answer 5

@eugesd recognize that \(\cos x=1-\frac{x^2}2+\frac{x^4}{4!}-\dots\). Similarly, \(\cos 5x=1-\frac{25x^2}2+\frac{625x^4}{4!}-\dots\). It follows that: $$\cos x-\cos 5x=1-1-\frac{x^2}2+\frac{25x^2}2+\frac{x^4}{4!}-\frac{625x^4}{4!}=\frac{24x^2}2-\frac{624x^4}{4!}+\dots$$Divide through by \(x^2\) to yield:$$\frac{\cos x-\cos 5x}{x^2}=\frac{24}2-\frac{624x^2}{4!}+\dots$$Take the limit as \(x\to0\):$$\lim_{x\to\infty}\frac{\cos x-\cos5x}{x^2}=\lim_{x\to\infty}\left(\frac{24}2-\frac{624x^2}{4!}+\dots\right)=\frac{24}2=12$$

Answer 6

oops I meant those limits to be \(x\to0\) not \(\infty\).

L'Hopital's rule provides the same result:$$\lim_{x\to0}\frac{\cos x-\cos5x}{x^2}=\lim_{x\to0}\frac{-\sin x+5\sin 5x}{2x}=\lim_{x\to0}\frac{-\cos x+25\cos x}{2}=\frac{-1+25}2=12$$

Answer 7

this one is just for fun, sub \(x=2y\) :)\[L=\lim_{x\to0}\frac{\cos x-\cos5x}{x^2}=\lim_{y\to0}\frac{\cos 2y-\cos10y}{4y^2}\]\[L=\lim_{y\to0}\frac{\cos 2y-\cos10y}{4y^2}=\lim_{y\to0}\frac{\cos^2y-\sin^2y-(\cos^25y-\sin^25y)}{4y^2}\]\[L=\lim_{y\to0}\frac{\cos^2y-\cos^25y-\sin^2y+\sin^25y}{4y^2}\]\[L=\frac{1}{4}\lim_{y\to0}\frac{\cos y-\cos5y}{y^2}(\cos y+\cos5y)-\frac{1}{4}\lim_{y\to0}\frac{\sin^2y}{y^2}+\frac{1}{4}\lim_{y\to0}\frac{\sin^25y}{y^2}\]\[L=\frac{L}{2}-\frac{1}{4}+\frac{25}{4}\]\[L=12 \]