I need some help with calculation of limits. I have a function $n(e^{it/\sqrt{m+n}}-1) + m(e^{-it/\sqrt{m+n}}-1) + \frac{m-n}{\sqrt{m+n}}it$
The solution says this converges to $-\frac{1}{2}t^2 \mathrm{~for~} n,m \rightarrow \infty$, but I still haven't figured out yet, how they did it. I separated this function into three parts, considering every summand individually.
For: $n(e^{it/\sqrt{m+n}}-1)$ I used the Euler's Formula and I got: $n(cos(\frac{t}{\sqrt{m+n}})+isin(\frac{t}{\sqrt{m+n}})-1)$.
Now if I look at the argument of sin and cos, they both tend to $0$ for $n,m \rightarrow \infty$, so I have left: $n(1+0-1)=0$
And now the same for the $m(e^{-it/\sqrt{m+n}}-1) \rightarrow 0$ But the last summand does not converge to $-\frac{1}{2}t^2$.
What did I do wrong? How do I usually handle this kind of limit?
My other idea was to substitute $u=\frac{t}{\sqrt{m+n}}$ but I cannot rearrange it to get to something like $\frac{1-cos(u )}{u}$ or $\frac{sin(u )}{u}$
Thank you!
In order to evaluate the limit, we must retain terms of second order in the expansion of the complex exponential terms.
Note that we can write
$$e^{\pm it/\sqrt{m+n}}-1= \pm it\,\frac{1}{\sqrt{m+n}}-\frac12 \frac{t^2}{m+n}+O\left(\frac{1}{(m+n)^{3/2}}\right)$$
Then, we have
$$\begin{align} n\left(e^{ it/\sqrt{m+n}}-1\right)+m\left(e^{- it/\sqrt{m+n}}-1\right)+it\frac{m-n}{\sqrt{m+n}}&=-\frac12 t^2+O\left(\frac{1}{\sqrt{m+n}}\right)\\\\ &\to -\frac12 t^2 \,\,\text{as}\,\,m,n\to \infty \end{align}$$