I need to prove this equation:
$$\lim_{r\to0}\left((1+\frac{r}{1})^1-(1+\frac{r}{n})^n\right)=\frac{1-n}{2n}r^2$$
Lecturer gave us a hint: $$\lim_{x\to0} \frac{\sin x}{x}=1$$
I was thinking about expanding $(1+r/n)^n$ with Taylor series, but that gave me nothing.
Do you have any ideas/suggestions? Thank you.
As given, the equation is not correct. The limit is zero, which is also the limit of the right side as $r \to 0$. What you have is the leading order (smallest power of $r$) term in the expansion. The hint you are given is correct, but I don't know what use it is. You are expected to be expanding $\left(1+\frac rn\right)^n$ using the binomial theorem. The first two terms will cancel with the stuff before the minus sign. The limit you are quoting will be the next term.
Almost as said in Ross Millikan's answer.
If you do not use the binomial theorem, consider $$A=\left(1+\frac{r}{n}\right)^n\implies \log(A)=n\log\left(1+\frac{r}{n}\right)$$ Since $\frac rn$ is small compared to $1$, use Taylor expansion (just as you though about it) $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)$$ Replace $x$ by $\frac rn$ to get $$\log(A)=n \times \left(\frac{r}{n}-\frac{r^2}{2 n^2}+\frac{r^3}{3 n^3}+O\left(r^4\right) \right)=r-\frac{r^2}{2 n}+\frac{r^3}{3 n^2}+O\left(r^4\right)$$ Now, use Taylor again using $$A=e^{\log(A)}=1+r+\frac{(n-1) }{2 n}r^2+\frac{\left(n^2-3 n+2\right) }{6 n^2}r^3+O\left(r^4\right)$$ This makes $$\left(1+\frac{r}{1}\right)^1-\left(1+\frac{r}{n}\right)^n=-\frac{(n-1) }{2 n}r^2-\frac{(n-2)(n-3) }{6 n^2}r^3+O\left(r^4\right)$$ If you limit the expansion to $O\left(r^3\right)$, you get your result.