Question

1. this one is difficult to start so i started it and see if you guys can do the last couple of steps.

lim_{x \to 1} \frac{5x^{2}-2x^{3}+2x - 5}{4x^{2}+6x - 10x^{3}}
lim_{x \to 1} \frac{-2x^{3}+5x^{2}+2x - 5}{-10x^{3}+4x^{2}+6x}
lim_{x \to 1} \frac{-x^{2}(2x - 5)+1(2x - 5)}{-2x(5x^{2}-2x - 3)}
lim_{x \to 1} \frac{(2x - 5)(1 - x^{2})}{-2x(5x + 3)(x - 1)}
lim_{x \to 1} \frac{(2x - 5)(x - 1)(x + 1)}{-2x(5x + 3)(x - 1)}

Explanation:

Step1: Simplify the rational - function

First, we have the limit $\lim_{x
ightarrow1}\frac{5x^{2}-2x^{3}+2x - 5}{4x^{2}+6x - 10x^{3}}$.
Factor the numerator and denominator.
For the numerator $5x^{2}-2x^{3}+2x - 5=-2x^{3}+5x^{2}+2x - 5=-x^{2}(2x - 5)+1(2x - 5)=(2x - 5)(1 - x^{2})=(2x - 5)(1 - x)(1 + x)$.
For the denominator $4x^{2}+6x - 10x^{3}=-10x^{3}+4x^{2}+6x=-2x(5x^{2}-2x - 3)=-2x(5x + 3)(x - 1)$.
So the limit becomes $\lim_{x
ightarrow1}\frac{(2x - 5)(1 - x)(1 + x)}{-2x(5x + 3)(x - 1)}$.

Step2: Cancel out the common factor

We can cancel out the common factor $(x - 1)$ (note that $x
ightarrow1$ but $x
eq1$).
$\lim_{x
ightarrow1}\frac{(2x - 5)(1 - x)(1 + x)}{-2x(5x + 3)(x - 1)}=\lim_{x
ightarrow1}\frac{-(2x - 5)(x - 1)(1 + x)}{-2x(5x + 3)(x - 1)}=\lim_{x
ightarrow1}\frac{(2x - 5)(1 + x)}{2x(5x + 3)}$.

Step3: Substitute $x = 1$

Substitute $x = 1$ into $\frac{(2x - 5)(1 + x)}{2x(5x + 3)}$.
When $x = 1$, we have $\frac{(2\times1 - 5)(1 + 1)}{2\times1\times(5\times1+3)}=\frac{(2 - 5)\times2}{2\times(5 + 3)}=\frac{-3\times2}{2\times8}=-\frac{3}{8}$.

Answer:

$-\frac{3}{8}$