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\(\displaystyle L=\lim_{t\to0}\frac{e^{2t}-1}{1-\cos(t)}\)

This is the indeterminate form \(\displaystyle \frac{0}{0}\), so application of L'Hôpital's rule yields:

\(\displaystyle L=\lim_{t\to0}\frac{2e^{2t}}{\sin(t)}=\frac{2}{0}\)

This is undefined, so we want to look at the one-sided limits:

\(\displaystyle \lim_{t\to0^{-}}\frac{2e^{2t}}{\sin(t)}=-\infty\)

\(\displaystyle \lim_{t\to0^{+}}\frac{2e^{2t}}{\sin(t)}=\infty\)

Since:

\(\displaystyle \lim_{t\to0^{-}}\frac{2e^{2t}}{\sin(t)}\ne \lim_{t\to0^{+}}\frac{2e^{2t}}{\sin(t)}\) we may conclude that the limit $L$ does not exist.