Oblicz granicę za pomocą reguły de L'Hospitala.... Question from @kamil2001i

kamil2001i @kamil2001i

June 2022 1 34 Report

Oblicz granicę za pomocą reguły de L'Hospitala

Odpowiedź:

$\lim_{x\to\alpha}\mathrm{tg}\bigg(\frac{\pi x}{2\alpha }\bigg)\big(e^{\mathrm{sin}(a)}- e^{\mathrm{sin}(x)}\big)=\frac{2\alpha \cdot e^{\mathrm{sin}(\alpha )}\cdot\mathrm{cos}(\alpha )}{\pi}$

Szczegółowe wyjaśnienie:

$\lim_{x\to\alpha}\mathrm{tg}\bigg(\frac{\pi x}{2\alpha }\bigg)\big(e^{\mathrm{sin}(a)}- e^{\mathrm{sin}(x)}\big)=\lim_{x\to\alpha}\mathrm{sin}\bigg(\frac{\pi x}{2\alpha }\bigg)\big(e^{\mathrm{sin}(a)}- e^{\mathrm{sin}(x)}\big)\cdot\frac{1}{\mathrm{cos}\bigg(\frac{\pi x}{2\alpha }\bigg)}$

Korzystamy z własności na iloczyn granic (Uwaga: obie granice muszą istnieć!). Spróbujmy zatem zapisać to w postaci iloczynu granic:

$\lim_{x\to\alpha }\mathrm{sin}\bigg(\frac{\pi x}{2\alpha}\bigg)\cdot \lim_{x\to\alpha}\frac{\big(e^{\mathrm{sin}(\alpha )}-e^{\mathrm{sin}(x)}\big)}{\mathrm{cos}\bigg(\frac{\pi x}{2\alpha}\bigg)}$

Pierwsza z granic oczywiście istnieje:

$\lim_{x\to\alpha }\mathrm{sin}\bigg(\frac{\pi x}{2\alpha}\bigg)=\mathrm{sin}\bigg(\frac{\pi}{2}\bigg)=1$

Przeanalizujmy drugą z granic:

$\lim_{x\to\alpha}\frac{\big(e^{\mathrm{sin}(\alpha )}-e^{\mathrm{sin}(x)}\big)}{\mathrm{cos}\bigg(\frac{\pi x}{2\alpha}\bigg)}=_H^{\big[\frac{0}{0}\big]}=\lim_{x\to\alpha}\frac{-e^{\mathrm{sin}(x)}\cdot\mathrm{cos}(x)}{\frac{-\pi\cdot \mathrm{sin}\big(\frac{\pi x}{2\alpha }\big) }{2\alpha }} =\lim_{x\to\alpha}\frac{2\alpha \cdot e^{\mathrm{sin}(x)}\cdot\mathrm{cos}(x)}{\pi\cdot \mathrm{sin}\big(\frac{\pi x}{2\alpha }\big)} =$

$=\frac{2\alpha \cdot e^{\mathrm{sin}(\alpha )}\cdot\mathrm{cos}(\alpha )}{\pi}\Longrightarrow\lim_{x\to\alpha}\frac{\big(e^{\mathrm{sin}(\alpha )}-e^{\mathrm{sin}(x)}\big)}{\mathrm{cos}\bigg(\frac{\pi x}{2\alpha}\bigg)} \ \ \exists}$

Zatem:

$\lim_{x\to\alpha}\mathrm{tg}\bigg(\frac{\pi x}{2\alpha }\bigg)\big(e^{\mathrm{sin}(a)}- e^{\mathrm{sin}(x)}\big)=1\cdot \frac{2\alpha \cdot e^{\mathrm{sin}(\alpha )}\cdot\mathrm{cos}(\alpha )}{\pi}=\frac{2\alpha \cdot e^{\mathrm{sin}(\alpha )}\cdot\mathrm{cos}(\alpha )}{\pi}$