Fórum de Matemática
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Como é que resolvo estes limites?
1)\(\lim_{x-0}\frac{ln(2x+1))}{e^x-1}\)
2)\(\lim_{ x \to 1} \; \frac{x^2-x}{2x*\ln x}\)

2 )

\(\lim_{ x \to 1} \; \frac{x^2-x}{2x*\ln x}\)


\(\lim_{ x \to 1} \; \frac{x-1}{2\ln x}\)


\(u=x-1 \;\; x=u+1 \;\;\;\;\; x \to 1 \;\; u \to 0\)


\(\lim_{ u \to 0} \; \frac{u}{2\ln (u+1)}\)


\(\frac{1}{2}*\lim_{ u \to 0} \; \frac{u}{\ln (u+1)}\)


\(\frac{1}{2}*\lim_{ u \to 0} \; \frac{1}{\frac{\ln (u+1)}{u}}\)


\(\frac{1}{2}*\lim_{ u \to 0} \; \frac{1}{\ln (u+1)^{\frac{1}{u}}\)


\(\frac{1}{2}*\left( \; \frac{\lim_{ u \to 0} \; 1}{\lim_{ u \to 0} \; \ln (u+1)^{\frac{1}{u}} \right)\)


\(\frac{1}{2}*\left( \; \frac{\lim_{ u \to 0} \; 1}{\ln \left( \lim_{ u \to 0} \; (u+1)^{\frac{1}{u} \right) \right)\)


\(\fbox{\fbox{\fbox{\frac{1}{2}}}}\)



PS: Verifique o primeiro limite.Por favor.

1) comece dividindo por x:


\(\lim_{ x \to 0 } \; \frac{\frac{\ln(2x+1)}{x}}{\frac{e^{x}-1}{x}}\)


\(\lim_{ x \to 0 } \; \frac{\ln(2x+1)}{x}\)


\(u=\ln(2x+1) \;\; \frac{e^{u}-1}{2}=x \;\;\;\;\; x \to 0 \;\; u \to 0\)


\(\lim_{ u \to 0} \; \frac{u}{ \frac{e^{u}-1}{2}}\)



\(\lim_{ u \to 0} \; \frac{1}{ \frac{e^{u}-1}{2u}}\)



\(\frac{\lim_{ u \to 0} \; 1}{ \lim_{ u \to 0} \; \frac{e^{u}-1}{2u}}\)


\(\frac{\lim_{ u \to 0} \; 1}{ \frac{1}{2}*\lim_{ u \to 0} \; \frac{e^{u}-1}{u}}\)


\(\frac{ \; 1}{ \frac{1}{2}}=\fbox{\fbox{2}}\)