Granica

[Ukasz12344]

Oblicz granice ln(4x+1/4x-7)^x przy x->oo

[panb]

\( \Lim_{x\to \infty } \ln \left( \frac{4x+1}{4x-7} \right)^x =\ln \left[ \Lim_{x\to \infty }\left( \frac{4x+1}{4x-7} \right)^x\right] \\
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\left( \frac{4x+1}{4x-7} \right)^x= \left( \frac{4x-7+8}{4x-7} \right)^x= \left( 1+ \frac{8}{4x-7} \right)^x= \left(1+ \frac{1}{ \frac{1}{2} x- \frac{7}{8}} \right)^x = \left[ \left(1+ \frac{1}{ \frac{1}{2} x- \frac{7}{8}} \right)^{\frac{1}{2} x- \frac{7}{8}}\right] ^2 \cdot \left(1+ \frac{1}{ \frac{1}{2} x- \frac{7}{8}}\right)^{ \frac{7}{4} } \)
Ponieważ:
\[ \Lim_{x\to \infty } \left[ \left(1+ \frac{1}{ \frac{1}{2} x- \frac{7}{8}} \right)^{\frac{1}{2} x- \frac{7}{8}}\right] ^2=e^2\\
\Lim_{x\to \infty } \left(1+ \frac{1}{ \frac{1}{2} x- \frac{7}{8}}\right)^{ \frac{7}{4} }=1 \]
więc

Odpowiedź: \(\Lim_{x\to \infty } \ln \left( \frac{4x+1}{4x-7} \right)^x =\ln e^2=2\)

[korki_fizyka]

Ukasz nałucz sie pisac w Latexie: https://forum.zadania.info/viewtopic.php?f=6&t=568