Question

lim x tends to 1, X4-3X2+2/X3-5X2+3x+1

Answer 1

hope it helps!!!!!!!!!!

Answer 2

$\lim_{x \to 1 } \dfrac{x^{4}-3x^{2}+2}{x^{3}-5x^{2}+3x+1}=\dfrac{1}{2}.$

Step-by-step explanation:

We have,

$\lim_{x \to 1 } \dfrac{x^{4}-3x^{2}+2}{x^{3}-5x^{2}+3x+1}$

Put x = 1, we get

$\dfrac{1^{4}-3(1)^{2}+2}{1^{3}-5(1)^{2}+3(1)+1}$ = $\dfrac{0}{0}$, form

Applying L'Hospital rule, we get

$\lim_{x \to 1 } \dfrac{4x^{3}-6x+0}{3x^{2}-10x+3+0}$

= $\lim_{x \to 1 } \dfrac{4x^{3}-6x}{3x^{2}-10x+3}$

Put x = 1, we get

$=\dfrac{4(1)^{3}-6(1)}{3(1)^{2}-10(1)+3}$

$=\dfrac{4-6}{6-10}$

$=\dfrac{-2}{-4}$

$=\dfrac{1}{2}$

Hence,$\lim_{x \to 1 } \dfrac{x^{4}-3x^{2}+2}{x^{3}-5x^{2}+3x+1}=\dfrac{1}{2}$.