How to prove that
lim
without using L'Hospital?
Using L'Hospital, it's quite easy. But without, I don't get this. I tried different approaches, for example writing e^{\sin(x)}=\sum\limits_{k=0}^\infty\frac{\sin(x)^k}{k!}
and
\sin(2x)=2\sin(x)\cos(x)
and get
\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}
but it seems to be unrewarding. How can I calculate the limit instead?
Any advice will be appreciated.
Answer
From the known limit
\lim\limits_{u\to 0}\frac{e^u-1}{u}=1, one gets
\lim\limits_{x\to 0}\frac{e^{\sin x}-1}{\sin(2x)}=\lim\limits_{x\to 0}\left(\frac{e^{\sin x}-1}{\sin x}\cdot\frac{\sin x}{\sin(2x)}\right)=\lim\limits_{x\to 0}\left(\frac{e^{\sin x}-1}{\sin x}\cdot\frac{1}{2\cos x}\right)=\color{red}{1}\cdot\frac{1}{2}.
No comments:
Post a Comment