lim _xgoes to 2(fracx^{4-4x^3+5x^2-4x+4}x^{4-4x^3+6x^2-8x+8})

Frequently Asked Questions

• Evaluate using transformations of ${\lim }_{x\to 2}\left(\frac{x^4-4x^3+5x^2-4x+4}{x^4-4x^3+6x^2-8x+8}\right)$

  

  $$\frac{5}{6}$$

• Evaluate using L'Hopital's rule of ${\lim }_{x\to 2}\left(\frac{x^4-4x^3+5x^2-4x+4}{x^4-4x^3+6x^2-8x+8}\right)$

  

  $$\frac{5}{6}$$