svp aidez moi a resoudre ce exercice dm. calculer lim(x²-4x+1)= x⇒-∞ lim(-x³+x+1)= x⇒-∞ lim 3x-1/2x x⇒-∞ lim x²-4x+1/x-1 x⇒-∞ lim 3x-1/2x x⇒+∞ lim x²-4x+1/x-1 x

lim(x²-4x+1)= lim(x²) = +∞
x⇒-∞             x⇒-∞

lim(-x³+x+1)= lim (-x³) = -(-∞) = +∞
x⇒-∞             x⇒-∞

lim 3x-1/2x= lim (3x/2x) = 3/2
x⇒-∞           x⇒-∞           

lim x²-4x+1/x-1 = lim (x²/x) = lim x = -∞
x⇒-∞                  x⇒-∞         x⇒-∞

lim 3x-1/2x = lim (3x/2x) = 3/2
x⇒+∞           x⇒+∞          

lim x²-4x+1/x-1 = lim (x²/x) = lim x = +∞
x⇒+∞                 x⇒+∞        x⇒+∞

[tex] \lim_{x \to -\infty} x^2-4x+1= \lim_{x \to -\infty} x^2=\infty\\\\ \lim_{x \to -\infty}-x^3+x+1= \lim_{x \to -\infty} -x^3=-\infty\\\\ \lim_{x \to -\infty} \frac{3x-1}{2x}= \lim_{x \to -\infty} \frac{3x}{2x}= \frac{3}{2}\\\\ \lim_{x \to -\infty} \frac{x^2-4x+1}{x-1}=\lim_{x \to -\infty} \frac{x^2}{x} = \lim_{x \to -\infty}x=-\infty\\\\ \lim_{x \to \infty} \frac{3x-1}{2x}=\lim_{x \to \infty} \frac{3x}{2x}=\lim_{x \to \infty} \frac{3}{2}= \frac{3}{2} \\\\ [/tex]

[tex]\lim_{x \to \infty} \frac{x^2-4x+1}{x-1}=\lim_{x \to \infty} \frac{x^2}{x}=\lim_{x \to \infty} x=\infty [/tex]