Question

Determine o limite 3x^3-5x^2+x+1÷2x^3-3x^2+1
x tende a 1​

Answer 1

$lim_{x⟶1}( \frac{3x {}^{3} - 5x {}^{2} + x + 1 }{2x {}^{3} - 3x {}^{2} + 1 } )$

$lim_{x⟶1}( \frac{3x {}^{3} - 3x {}^{2} - 2x {}^{2} + 2x - x + 1}{2x {}^{3} - 2x {}^{2} - x {}^{2} + x - x + 1} )$

$lim_{x⟶1}( \frac{3x {}^{2} \: . \: (x - 1) - 2x \: . \: (x - 1) - (x - 1) }{2x {}^{2} \: . \: (x - 1) - x \: . \: (x - 1) - (x - 1)} )$

$lim_{x⟶1}( \frac{(x - 1) \: . \: (3x {}^{2} - 2x - 1)}{(x - 1) \: . \: (2x {}^{2} - x - 1} )$

$lim_{x⟶1}( \frac{3x {}^{2} + x - 3x - 1}{2x {}^{2} + x - 2x - 1 }$

$lim_{x⟶1}( \frac{x \: . \: (3x + 1) - (3x + 1)}{x \: . \: (2x + 1) - (2x + 1)} )$

$lim_{x⟶1}( \frac{(3x + 1) \: . \: (x - 1)}{(2x + 1) \: . \: (x - 1)} )$

$lim_{x⟶1} (\frac{3x + 1}{2x + 1} )$

$\frac{3 \:. \:1 + 1 }{2 \: . \: 1 + 1}$

$\frac{3 + 1}{2 + 1}$

$\boxed{\boxed{\boxed{ \frac{4}{3} }}}$

att. yrz