Friday, 18 May 2018

calculus - Should L'Hopital's Rule be used for this limit?



Question



limx0(sinxx)11cosx



Attempt




This limit is equal to,



\lim_{x \to 0} \exp\left({\dfrac{1}{1 - \cos x}\ln \left(\dfrac{\sin x}{x}\right)}\right).



I hope to solve this by focusing on the limit,



\lim_{x \to 0} {\dfrac{1}{1 - \cos x}\ln \left(\dfrac{\sin x}{x}\right)}.



Which is indeterminate of the form "\frac{0}{0}". However, repeated application of L'Hopital's Rule seems to lead to an endless spiral of trigonometric terms, but wolfram Alpha says the limit I am focusing on exists.




Is there another way to solve this?


Answer



It should be used. There may be some little mistakes in your computation.



\begin{align*} \lim_{x\rightarrow0}\frac{1}{1-\cos x}\ln\frac{\sin x}{x} &=\lim_{x\rightarrow0}\frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{\sin x}\\ &=\lim_{x\rightarrow0}\frac{x\cos x-\sin x}{x\sin^{2}x}\\ &=\lim_{x\rightarrow0}\frac{-x\sin x}{\sin^{2}x+2x\sin x\cos x}\\ &=-\lim_{x\rightarrow0}\frac{1}{\frac{\sin x}{x}+2\cos x}\\ &=-\frac{1}{3} \end{align*}


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