tim gioi han

N

nguyenbahiep1

g(x)= (\sqrt{x^2+x+2}- x^2- x)/(x-1) khi x\Rightarrow 1

[laTEX]L = \lim_{x \to 1} \frac{\sqrt{x^2+x+2}- 2}{x-1} - \lim_{x \to 1} \frac{x^2+x- 2}{x-1} \\ \\ \lim_{x \to 1} \frac{x^2+x-2}{(x-1)(\sqrt{x^2+x+2}+2)} - \lim_{x \to 1} \frac{(x-1)(x+2)}{x-1} \\ \\ \lim_{x \to 1} \frac{x+2}{\sqrt{x^2+x+2}+2} - \lim_{x \to 1} (x+2) \\ \\ L = \frac{3}{4} - 3 = - \frac{9}{4}[/laTEX]
 
N

noinhobinhyen

$\lim_{x \to +\infty} \dfrac{2-x^2}{\sqrt{x^2+1}-x}$

Ta có $\lim_{x \to +\infty} (2-x^2)= -\infty$

$\lim_{x \to +\infty} (\sqrt{x^2+1}-x) = 0^+$

$\Rightarrow \lim_{x \to +\infty} \dfrac{2-x^2}{\sqrt{x^2+1}-x} = -\infty$
 
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