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Vypo�t�t� limitu:

$$\lim\limits_{x \to 0} {e^{ax}-e^{bx} \over \sin ax - \sin bx}$$

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Dan� v�raz uprav�me tak, abychom mohli pou��t vzorce $\lim\limits_{x \to 0} {e^x - 1 \over x} = 1$  a $\lim\limits_{x \to 0} {\sin x \over x} = 1$:

$$\lim\limits_{x \to 0} {e^{ax}-e^{bx} \over \sin ax - \sin bx}=\lim\limits_{x \to 0} {e^{bx}(e^{(a-b)x}-1) \over 2 \cos \left({a+b \over 2} x\right) \sin {\left({a-b \over 2} x\right)}} = \lim\limits_{x \to 0} {(a-b)x e^{bx} \over 2 \cos \left({a+b \over 2} x\right) \sin {\left({a-b \over 2} x\right)}} {e^{(a-b)x} -1 \over (a-b)x}  = 1
$$

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