Odpowiedź:
[tex]$\lim_{x\to\alpha}\mathrm{tg}\bigg(\frac{\pi x}{2\alpha }\bigg)\big(e^{\mathrm{sin}(a)}- e^{\mathrm{sin}(x)}\big)=\frac{2\alpha \cdot e^{\mathrm{sin}(\alpha )}\cdot\mathrm{cos}(\alpha )}{\pi}$[/tex]
Szczegółowe wyjaśnienie:
[tex]$\lim_{x\to\alpha}\mathrm{tg}\bigg(\frac{\pi x}{2\alpha }\bigg)\big(e^{\mathrm{sin}(a)}- e^{\mathrm{sin}(x)}\big)=\lim_{x\to\alpha}\mathrm{sin}\bigg(\frac{\pi x}{2\alpha }\bigg)\big(e^{\mathrm{sin}(a)}- e^{\mathrm{sin}(x)}\big)\cdot\frac{1}{\mathrm{cos}\bigg(\frac{\pi x}{2\alpha }\bigg)} $[/tex]
Korzystamy z własności na iloczyn granic (Uwaga: obie granice muszą istnieć!). Spróbujmy zatem zapisać to w postaci iloczynu granic:
[tex]$\lim_{x\to\alpha }\mathrm{sin}\bigg(\frac{\pi x}{2\alpha}\bigg)\cdot \lim_{x\to\alpha}\frac{\big(e^{\mathrm{sin}(\alpha )}-e^{\mathrm{sin}(x)}\big)}{\mathrm{cos}\bigg(\frac{\pi x}{2\alpha}\bigg)}$[/tex]
Pierwsza z granic oczywiście istnieje:
[tex]$\lim_{x\to\alpha }\mathrm{sin}\bigg(\frac{\pi x}{2\alpha}\bigg)=\mathrm{sin}\bigg(\frac{\pi}{2}\bigg)=1 $[/tex]
Przeanalizujmy drugą z granic:
[tex]$\lim_{x\to\alpha}\frac{\big(e^{\mathrm{sin}(\alpha )}-e^{\mathrm{sin}(x)}\big)}{\mathrm{cos}\bigg(\frac{\pi x}{2\alpha}\bigg)}=_H^{\big[\frac{0}{0}\big]}=\lim_{x\to\alpha}\frac{-e^{\mathrm{sin}(x)}\cdot\mathrm{cos}(x)}{\frac{-\pi\cdot \mathrm{sin}\big(\frac{\pi x}{2\alpha }\big) }{2\alpha }} =\lim_{x\to\alpha}\frac{2\alpha \cdot e^{\mathrm{sin}(x)}\cdot\mathrm{cos}(x)}{\pi\cdot \mathrm{sin}\big(\frac{\pi x}{2\alpha }\big)} =$[/tex]
[tex]$=\frac{2\alpha \cdot e^{\mathrm{sin}(\alpha )}\cdot\mathrm{cos}(\alpha )}{\pi}\Longrightarrow\lim_{x\to\alpha}\frac{\big(e^{\mathrm{sin}(\alpha )}-e^{\mathrm{sin}(x)}\big)}{\mathrm{cos}\bigg(\frac{\pi x}{2\alpha}\bigg)} \ \ \exists}$[/tex]
Zatem:
[tex]$\lim_{x\to\alpha}\mathrm{tg}\bigg(\frac{\pi x}{2\alpha }\bigg)\big(e^{\mathrm{sin}(a)}- e^{\mathrm{sin}(x)}\big)=1\cdot \frac{2\alpha \cdot e^{\mathrm{sin}(\alpha )}\cdot\mathrm{cos}(\alpha )}{\pi}=\frac{2\alpha \cdot e^{\mathrm{sin}(\alpha )}\cdot\mathrm{cos}(\alpha )}{\pi}$[/tex]
