calcular estes limites:
![\lim_{x \to \infty} [ln(2x+1)-ln(x+3)]](https://tex.z-dn.net/?f=+%5Clim_%7Bx+%5Cto+%5Cinfty%7D+%5Bln%282x%2B1%29-ln%28x%2B3%29%5D "\lim_{x \to \infty} [ln(2x+1)-ln(x+3)]")
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Primeira:
![\lim_{x\to\infty}\ln\frac{x}{x+1}=\\\\\ln[\lim_{x\to\infty}\frac{x}{x(1+\frac{1}{x})}]=\\\\\ln[\frac{1}{1+0}]=\\\\\ln1=\\\\\boxed{0}](https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cln%5Cfrac%7Bx%7D%7Bx%2B1%7D%3D%5C%5C%5C%5C%5Cln%5B%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7Bx%7D%7Bx%281%2B%5Cfrac%7B1%7D%7Bx%7D%29%7D%5D%3D%5C%5C%5C%5C%5Cln%5B%5Cfrac%7B1%7D%7B1%2B0%7D%5D%3D%5C%5C%5C%5C%5Cln1%3D%5C%5C%5C%5C%5Cboxed%7B0%7D "\lim_{x\to\infty}\ln\frac{x}{x+1}=\\\\\ln[\lim_{x\to\infty}\frac{x}{x(1+\frac{1}{x})}]=\\\\\ln[\frac{1}{1+0}]=\\\\\ln1=\\\\\boxed{0}")
Segunda:
![\lim_{x\to\infty}\ln\frac{2x+1}{x+3}=\\\\\ln[\lim_{x\to\infty}\frac{x(2+\frac{1}{x})}{x(1+\frac{3}{x})}]=\\\\\ln[\lim_{x\to\infty}\frac{2+\frac{1}{x}}{1+\frac{3}{x}}]=\\\\\ln(\frac{2+0}{1+0})=\\\\\boxed{\ln2}](https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cln%5Cfrac%7B2x%2B1%7D%7Bx%2B3%7D%3D%5C%5C%5C%5C%5Cln%5B%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7Bx%282%2B%5Cfrac%7B1%7D%7Bx%7D%29%7D%7Bx%281%2B%5Cfrac%7B3%7D%7Bx%7D%29%7D%5D%3D%5C%5C%5C%5C%5Cln%5B%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B2%2B%5Cfrac%7B1%7D%7Bx%7D%7D%7B1%2B%5Cfrac%7B3%7D%7Bx%7D%7D%5D%3D%5C%5C%5C%5C%5Cln%28%5Cfrac%7B2%2B0%7D%7B1%2B0%7D%29%3D%5C%5C%5C%5C%5Cboxed%7B%5Cln2%7D "\lim_{x\to\infty}\ln\frac{2x+1}{x+3}=\\\\\ln[\lim_{x\to\infty}\frac{x(2+\frac{1}{x})}{x(1+\frac{3}{x})}]=\\\\\ln[\lim_{x\to\infty}\frac{2+\frac{1}{x}}{1+\frac{3}{x}}]=\\\\\ln(\frac{2+0}{1+0})=\\\\\boxed{\ln2}")
Segunda:
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