Saturday, 12 November 2016

limits - find: limnrightarrowinftyfracsinleft(x+frac1nright)sinleft(xright)sinleft(x+frac1nright)



Find: lim as x\in\mathbb{R}
My progress:




\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=\lim_{n\longrightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)}{\sin\left(x+\frac{1}{n}\right)}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=



\lim_{n\longrightarrow\infty} \ \ {1}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=1-\lim_{n\longrightarrow\infty}\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} at this point I got stuck.



I can't evaluate the Taylor series of \sin(x+\frac{1}{n}) because n is not fixed



(even if we'll suppose that there exist some \epsilon>0 and there exists N\in\mathbb{N}:\forall n\geq N s.t:
-\epsilon<\frac{1}{n}<\epsilon
it doesn't seem like a formal argument to me)




(I might be very wrong - it's only my intuition).



Also trying to apply L'Hopital's rule for this expression isn't much helpful.


Answer



If \sin(x) = 0 then
\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} = 1 \to 1
for n \to \infty, otherwise \sin\left(x+\frac{1}{n}\right) \to \sin(x) \ne 0 and therefore

\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} \to \frac{\sin(x) - \sin(x)}{\sin(x)} = 0 \, .



So there is no need to use L'Hospital's rule in the case \sin(x) = 0,
but doing so would give the same result:
\lim_{n \to \infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} = \lim_{h \to 0}\frac{\sin\left(x+h\right)-\sin\left(x\right)}{\sin\left(x+h\right)} \stackrel{\text{(H)}}{=} \lim_{h \to 0} \frac{\cos(x+h)}{\cos(x+h)} = 1 \, .


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