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Vypo�t�te limitu:

$$\lim\limits_{x \to 0} {(e^x + x)^{1\over x}}$$


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Jde o limitu typu $1^\infty$. Polo��me-li  $y=(e^x + x)^{1\over x}$ a logaritmujeme-li ob� strany rovnice, dostaneme:

$$\text{ln}\  y = {1 \over x} \text{ln}\ (e^x+x)$$

Pak po p�echodu k limit� dostaneme:

$$\lim\limits_{x \to 0} \text{ln}\  y = \lim\limits_{x \to 0} {\text{ln}\ {(e^x + x)^{1\over x}} \over x}= \lim\limits_{x \to 0} {{1 \over (e^x + x)}(e^x+1) \over 1}=2$$

Je tedy $\lim\limits_{x \to 0} \text{ln}\ y = 2$ a $\lim\limits_{x \to 0} y  = e^2$.


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