Question

Cat da la aceasta limita (cazul e 00-00)

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Answer 1

$l = \lim\limits_{x\to \infty}\left(\sqrt{x^2+3x+1}-x\right) \\ \\ =\lim\limits_{x\to \infty}\left(\sqrt{(x+\frac{3}{2})^2-\frac{9}{4}}-x\right)\\ \\\bullet\,\, x+\frac{3}{2} -\text{asimptota oblica la}+\infty\text{ pentru }\sqrt{x^2+3x+1}\\\\ \Rightarrow \sqrt{x^2+3x+1} \approx x+\frac{3}{2},\quad x\to \infty\\ \\ l=\lim\limits_{x\to \infty}\left[(x+\frac{3}{2}) - x\right]\\ \\ \Rightarrow \boxed{l = \dfrac{3}{2}}$

Answer 2

Răspuns:

Consideri expresia de la  limita ca fractie cu numitorul 1 si o amplifici cu

√(x²+3x+1)+x

Obtii

lim(√(x²+3x+1)-x)(√(x²+3x+1+x)/(√(x²+3x+1)+x)=

lim[x²+3x+1-x²]/(√(x²+3x+1)+x)=

lim(3x+1)/√x²(1+3/x+1/x²)+x)=

lim(3x+1)/(x√(1+3/x+1/x²+x)=

lim(3x+1)/x(1+3/x+1/x²)+1)=3/2    fiindca

3/x→0; 1/x²→0 deci radicalul tinde   la   1   si cu   1  din afara   radicalului fac 2

Explicație pas cu pas: