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Section 1.1 Détermination par l'algèbre

Déterminez les limites suivantes à l'aide de l'algèbre.

  1. \(\displaystyle\lim_{x\rightarrow -1}\frac{x+1}{x^2+2x+1}\)

  2. \(\displaystyle\lim_{x\rightarrow -1}\frac{2x^4+4x^2+2x^3+3x-1}{-2x^2+3x+5}\)

  3. \(\displaystyle\lim_{x\rightarrow 4}\frac{4-x}{\sqrt{4x}-4}\)

  4. \(\displaystyle\lim_{x\rightarrow 8}\frac{\frac{1}{3}-\frac{x}{24}}{x^2-64}\)

  5. \(\displaystyle\lim_{x\rightarrow +\infty}\frac{7x^2+2x^2-1}{5-28x^4}\)

  6. \(\displaystyle\lim_{x\rightarrow -\infty}\frac{x^2+4x-12}{2x^3-7x^2+7x-2}\)

  7. \(\displaystyle\lim_{x\rightarrow 1^+}\frac{3}{(x-1)^2}-\frac{2}{x-1}\)

  8. \(\displaystyle\lim_{x\rightarrow +\infty}\frac{\sqrt{4x^2-1}}{x+2}\)

  9. \(\displaystyle\lim_{x\rightarrow -\infty}\frac{12x+7}{\sqrt{9x^2+1}}\)

  10. \(\displaystyle\lim_{x\rightarrow 3}\frac{x^2-5x+6}{|x-3|}\)

  11. \(\displaystyle\lim_{x\rightarrow 3}\frac{x-3}{\sqrt{x+6}}\)

Réponse

  1. \(\displaystyle\lim_{x\rightarrow -1^-}\frac{x+1}{x^2+2x+1}=-\infty\) et \(\displaystyle\lim_{x\rightarrow -1^+}\frac{x+1}{x^2+2x+1}=+\infty\)

  2. \(\displaystyle -1\)

  3. \(\displaystyle -2\)

  4. \(\displaystyle -\frac{1}{384}\)

  5. \(\displaystyle -\frac{1}{4}\)

  6. \(\displaystyle 0\)

  7. \(\displaystyle +\infty\)

  8. \(\displaystyle 2\)

  9. \(\displaystyle -4\)

  10. \(\displaystyle\lim_{x\rightarrow 3^-}\frac{x^2-5x+6}{|x-3|}=-1\) et \(\displaystyle\lim_{x\rightarrow 3^+}\frac{x^2-5x+6}{|x-3|}=1\)

  11. \(\displaystyle 0\)