Granica z logarytmem naturalnym

[poetaopole]

\(\lim_{x \to 0} \frac{\ln(1+3x)}{x}\). Na podstawie rachunku granic, czyli bez reguły H.
A potem (ciągle bez reguły de Hospitala):
\(\lim_{ x\to 0} \sqrt[x]{1-2x}\).
\(\lim_{ x\to \infty }x( \ln(x+1)-lnx)\)
\(\lim_{x \to0 } \frac{\ln(x+3)-ln3}{x}\)

[eresh]

\(\Lim_{x\to 0}\frac{\ln (1+3x)}{x}=\Lim_{x\to 0}(\frac{\ln (1+3x)}{3x}\cdot 3)=3\Lim_{x\to 0}\frac{\ln (1+3x)}{3x}=3\cdot 1=3\)

[radagast]

A "od pieca" będzie tak:
\(e= \Lim_{n\to \infty } \left(1+ \frac{1}{n} \right)^n \So \\
1=\Lim_{n\to \infty } \ln \left(1+ \frac{1}{n} \right)^n= \Lim_{n\to \infty } n\ln \left(1+ \frac{1}{n} \right)=\Lim_{n\to \infty } \frac{ \ln \left(1+ \frac{1}{n} \right) }{ \frac{1}{n} }=\Lim_{u\to 0} \frac{ \ln \left(1+u \right) }{ u }= \Lim_{x\to 0} \frac{ \ln \left(1+3x \right) }{ 3x } \So\\
3= \Lim_{u\to 0} \frac{ \ln \left(1+3x \right) }{ x }\)

[eresh]

\(\lim_{ x\to 0} \sqrt[x]{1-2x}\).

\(\Lim_{x\to 0}\sqrt[x]{1-2x}=\Lim_{x\to 0}(1-2x)^{\frac{1}{x}}=\Lim_{x\to 0}(1-2x)^{(\frac{1}{-2x}\cdot (-2))}=e^{-2}\)

[eresh]

\(\lim_{ x\to \infty }x( \ln(x+1)-lnx)\)

\(\Lim_{x\to \infty}(x\ln \frac{x+1}{x})=\Lim_{x\to \infty}(x\ln(1+\frac{1}{x})=\Lim_{x\to 0}\frac{1}{x}\cdot\ln (1+x)=1\)

[eresh]

\(\lim_{x \to0 } \frac{\ln(x+3)-ln3}{x}\)

\(\Lim_{x\to 0}\frac{\ln\frac{x+3}{3}}{x}=\Lim_{x\to 0}\frac{1}{x}\ln (1+\frac{3}{x})=\Lim_{x\to 0}(\cdot\frac{1}{3}\frac{3}{x}\cdot \ln (1+\frac{3}{x}))=\frac{1}{3}\cdot 1=\frac{1}{3}\)

[Galen]

\(\lim_{ x\to \infty }x( \ln(x+1)-lnx)\)

\(\Lim_{x\to \infty}(x\ln \frac{x+1}{x})=\Lim_{x\to \infty}(x\ln(1+\frac{1}{x})=\Lim_{x\to 0}\frac{1}{x}\cdot\ln (1+x)=1\)

Można jeszcze tak;
\(\Lim_{x\to \infty }xln \frac{x+1}{x}= \Lim_{x\to \infty}(1+ \frac{1}{x})^x =ln e=1\)