Question

Limita pentru x tinde la 0 din x^2/(x-ln(x+1))

Answer 1

Răspuns:

$2$

Explicație pas cu pas:

$\lim_{n \to \ 0} \frac{x^2}{x-ln(x+1)}=Suntem~in~cazul~\frac{0}{0} ~si~aplicam~L'Hopital= \lim_{n \to \ 0} \frac{2x}{1-\frac{1}{x+1}}=Facem~calculele~la~numitor=\lim_{n \to \ 0} \frac{2x}{\frac{x+1}{x+1} -\frac{1}{x+1}}=\lim_{n \to \ 0} \frac{2x}{\frac{x+1-1}{x+1}}=\lim_{n \to \ 0} \frac{2x}{\frac{x}{x+1}}=\lim_{n \to \ 0} 2x*\frac{x+1}{x} =Simplificam~un~x=\lim_{n \to \ 0} 2*(x+1)=Trecem~la~limita=2*(0+1)=2$

Answer 2

$_{x \to o}^{lim}\ \frac{x^2}{x-ln(x+1)}=\frac{0}0\implies\ derivare\ prin\ l'Hospital\\\\ \implies_{x \to o}^{lim}\ \frac{(x^2)'}{[x-ln(x+1)]'}=\ _{x \to 0}^{lim}\ \frac{2x}{x'-[ln(x+1)]'}=\ _{x \to 0}^{lim}\ \frac{2x}{1-\frac{1}{x+1}*(x+1)'}=\ _{x \to 0}^{lim}\ \frac{2x}{1-\frac{1}{x+1}}=\\\\=\ _{x \to 0}^{lim}\ \frac{2x}{\frac{x+1-1}{x+1}}=\ _{x \to 0}^{lim}\ \frac{2x}{\frac{x}{x+1}}=\ _{x \to 0}^{lim}\ (2x\ {\cdot}\ \frac{x+1}{x})=\ _{x \to 0}^{lim}\ (2x+2)=2$