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Oblicz granicę:
a). \( \Lim_{x\to - \infty } \frac{\ln(1+2^x)}{\ln(1+4^x)} \)
b). \(\Lim_{x\to 0 } \frac{e^{ \alpha x}-\cos( \alpha x)}{e^{ \beta x}-\cos( \beta x)} \)

Filip25 pisze: ↑19 sty 2023, 16:11 Oblicz granicę:
a). \( \Lim_{x\to - \infty } \frac{ln(1+2^x)}{ln(1+4^x)} \)

\(\Lim_{x\to - \infty } \frac{\ln(1+2^x)}{\ln(1+4^x)}=\left[{0\over0}\right]\nad{H}{=}\Lim_{x\to-\infty}\dfrac{2^x\ln2(1+4^x)}{(1+2^x)4^x\ln4}=\Lim_{x\to-\infty}{1\over2}\cdot\dfrac{1+4^x}{(1+2^x)2^x}=\left[{1\over2}\cdot\frac{1+0}{(1+0)\cdot0^+}\right]=+\infty\)

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Filip25 pisze: ↑19 sty 2023, 16:11 Oblicz granicę:
b). \(\Lim_{x\to 0 } \frac{e^{ \alpha x}-cos( \alpha x)}{e^{ \beta x}-cos( \beta x)} \)

\(\Lim_{x\to 0 } \dfrac{e^{ \alpha x}-\cos( \alpha x)}{e^{ \beta x}-\cos( \beta x)}=\left[{0\over0}\right]\nad{H}{=}\Lim_{x\to 0 } \dfrac{\alpha e^{ \alpha x}+\alpha\sin( \alpha x)}{\beta e^{ \beta x}+\beta\sin( \beta x)}={\alpha\over\beta}\)

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