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 \, .
No comments:
Post a Comment