granica
: 23 sie 2007, o 19:04
oblicz granice
\(\displaystyle{ \lim_{x\to } \frac{\sin \frac{1}{x}}{1-e^\frac{1}{x}}}\)
\(\displaystyle{ \lim_{x\to } \frac{\sin \frac{1}{x}}{1-e^\frac{1}{x}}}\)
\(\displaystyle{ t=\frac{1}{x}\\
\lim\limits_{t\to 0}\frac{\sin{t}}{1-e^{t}}=^{H}=\lim\limits_{t\to 0}\frac{\cos{t}}{-e^{t}}=\frac{1}{-1}=-1}\)