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[aaleksaandraa]

\(\Lim_ {x\to 0} \frac{ln(2x+1)}{\sin(5x)}\)
\(\Limn \frac{xlnx}{x^2+1}\)

[eresh]

\(\Lim_ {x\to 0} \frac{ln(2x+1)}{\sin(5x)}\)

\(\Lim_{x\to 0}\frac{\ln (2x+1)}{\sin 5x}=^H\Lim_{x\to 0}\frac{\frac{2}{2x+1}}{5\cos x}=\frac{2}{5}\)

[panb]

\(\Lim_ {x\to 0} \frac{ln(2x+1)}{\sin(5x)}\)
\(\Limn \frac{xlnx}{x^2+1}\)

Albo bez l'Hospitala:
\(\Lim_ {x\to 0} \frac{ln(2x+1)}{\sin(5x)}= \Lim_ {x\to 0} \frac{ \frac{\ln(2x+1)}{2x}\cdot 2x }{ \frac{\sin5x}{5x}\cdot 5x }= \frac{1\cdot2}{1\cdot5} = \frac{2}{5} \)

[eresh]

\(\Limn \frac{xlnx}{x^2+1}\)

\(\Lim_{x\to \infty}\frac{x\ln x}{x^2+1}=\Lim_{x\to \infty}\frac{\ln x}{\frac{x^2+1}{x}}=^H\Lim_{x\to \infty}\frac{\frac{1}{x}}{\frac{2x\cdot x-x^2-1}{x^2}}=\Lim_{x\to \infty}(\frac{1}{x}\cdot\frac{x^2}{x^2-1})=\Lim_{x\to\infty}\frac{x}{x^2-1}=0\)

[panb]

\(\Limn \frac{xlnx}{x^2+1}\)

Albo bez ...
\(\Limn \frac{xlnx}{x^2+1}=\Limn \frac{ \frac{\ln x}{x} }{1+ \frac{1}{x^2} }= \frac{0}{1}=0 \)
(n, czy x tu ma być?)