Compute lim
What I have: \forall\ x\geq 0\ :\ x- \frac{x^2}{2}\leq \log(1+x)\leq x.
Apply to get that the limit equals 1 for any real number k.
Is this correct? Are there any other proofs?
Answer
Yes it works, here's another proof using a little more sofisticate tool (in this case unnecessary, but sometimes more useful).
By Stolz-Cesaro if (x_n) is a positive sequence and
\lim_n \dfrac{x_{n+1}}{x_n} = l
then
\lim_n \sqrt[n]{x_n} = l.
Taking as (x_n) the sequence you defined, an easy calculation shows that \dfrac{x_{n+1}}{x_n} \rightarrow 1,
therefore the thesis.
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