Calculate the limit with l'hospital's rule 
Calculate $$\lim_{\alpha\to 0}{-\exp(-\frac{\alpha t}{2m})\frac{F}{\alpha}\sqrt \frac mk\sin(\frac{t\sqrt{4km-\alpha^2}}{2m})+\frac F\alpha \sqrt{\frac mk}\sin{(t\sqrt{\frac km})}}$$

Here is what I tried: 
$$\lim_{\alpha\to 0}{-\exp(-\frac{\alpha t}{2m})\frac{F}{\alpha}\sqrt \frac mk\sin(\frac{t\sqrt{4km-\alpha^2}}{2m})+\frac F\alpha \sqrt{\frac mk}\sin{(t\sqrt{\frac km})}}$$
$$=\lim_{\alpha\to 0}\frac{F}{\alpha}\sqrt \frac mk({-\exp(-\frac{\alpha t}{2m})\sin(\frac{t\sqrt{4km-\alpha^2}}{2m})+\sin{(t\sqrt{\frac km}))}}$$
Applying l'hospital's rule, we get
$$=\lim_{\alpha\to 0}F\sqrt \frac mk({\frac{\alpha}{2m}\exp(-\frac{\alpha t}{2m})\sin(\frac{t\sqrt{4km-\alpha^2}}{2m})-\exp{(-\frac{\alpha t} {2m})}\cos{(\frac{t\sqrt{4km-\alpha^2}}{2m})}\frac t{2m}}\frac{-\alpha}{\sqrt{4km-\alpha^2}}$$
Then since the expression is a continuous function about $\alpha$, we plug in $\alpha=0$ and have that the limit vanishes. But this can be shown not to be the case.
 A: To find the limit without using L'Hospital's Rule, we will expand terms in powers of $\alpha$.  To that end, we note the following hold:
$$\begin{align}
\exp\left(-\frac{\alpha t}{2m}\right)&=1-\frac{\alpha t}{2m}+O(\alpha^2)\\\\
\sin \left(\frac{t\sqrt{4km-\alpha^2}}{2m} \,\, \right)&=\sin \left(t\sqrt{\frac{k}{m}}  \,\,\right)+O(\alpha^2)
\end{align}$$
Thus, 
$$\begin{align}
&-\exp\left(-\frac{\alpha t}{2m}\right)\,\frac{F}{\alpha} \sqrt{\frac{m}{k}}\,\,\sin \left(\frac{t\sqrt{4km-\alpha^2}}{2m} \,\, \right)+\frac{F}{\alpha}\sqrt{\frac{m}{k}}\sin \left(t\sqrt{\frac{k}{m}}  \,\,\right)\\\\
&=\frac{F}{\alpha}\sqrt{\frac{m}{k}}\left(-\exp\left(-\frac{\alpha t}{2m}\right)\,\,\,\sin \left(\frac{t\sqrt{4km-\alpha^2}}{2m} \,\, \right)+\sin \left(t\sqrt{\frac{k}{m}}  \,\,\right)\right)\\\\
&=\frac{F}{\alpha}\sqrt{\frac{m}{k}}\,\,\left(\sin \left(t\sqrt{\frac{k}{m}}  \,\,\right)\,\,\frac{\alpha t}{2m}+O(\alpha^2)\right)\\\\
&=\frac{F}{2k}\left(t\sqrt{\frac{k}{m}}\right)\sin \left(t\sqrt{\frac{k}{m}}  \,\,\right)+O(\alpha) \\\\
&\to \frac{F}{2k}\left(t\sqrt{\frac{k}{m}}\right)\sin \left(t\sqrt{\frac{k}{m}}  \,\,\right)\,\,\,\, \text{as} \,\,\,\,\,\alpha \to 0
\end{align}$$
