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Sa se arate ca functia \( f(x)=\frac {1-\cot x}{e^x} \) nu are limita la \( \infty \) .

\( f(x)=\frac{\sin{x}-\cos{x}}{\sin{x}e^x} \)
Luam doua subsiruri care tind la \( \infty \):

\( x_n=(2n+\frac{1}{2})\pi \)

\( y_n=(2n+\frac{1}{e^{2n\pi}})\pi \)

Avem \( \lim\limits_{n\rightarrow\infty}f(x_n)=\lim\limits_{n\rightarrow\infty}\frac{\sin{\frac{\pi}{2}}-\cos{\frac{\pi}{2}}}{\sin{\frac{\pi}{2}}e^{(2n+\frac{1}{2})\pi}}=0 \)

\( \lim\limits_{n\rightarrow\infty}f(y_n)=
\lim\limits_{n\rightarrow\infty}
\frac
{\sin{\frac{\pi}{e^{2n\pi}}}-\cos{\frac{\pi}{e^{2n\pi}}}}
{\sin{\frac{\pi}{e^{2n\pi}}}e^{(2n+\frac{1}{e^{2n\pi}})\pi}}=
\lim\limits_{n\rightarrow\infty}\frac{
\sin{\frac{\pi}{e^{2n\pi}}}-\cos{\frac{\pi}{e^{2n\pi}}}}
{\frac{\sin{\frac{\pi}{e^{2n\pi}}}}{\frac{\pi}{e^{2n\pi}}}
e^{\frac{\pi}{e^{2n\pi}}}\pi}=-\frac{1}{\pi} \)

Si deci, obtinem limite diferite.