Question

$\lim\limits_{x \to 1} \frac{5x^2 - 2x^3 + 2x - 5}{4x^2 + 6x - 10x^3}$

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(5x^{2}-2x^{3}+2x - 5\):
Rearrange terms: \(-2x^{3}+5x^{2}+2x - 5=-2x^{2}(x - \frac{5}{2})+2(x - \frac{5}{2})\)? Wait, better to group:
\((5x^{2}-5)+(-2x^{3}+2x)=5(x^{2}-1)-2x(x^{2}-1)=(x^{2}-1)(5 - 2x)=(x - 1)(x + 1)(5 - 2x)\)

Denominator \(4x^{2}+6x-10x^{3}=-10x^{3}+4x^{2}+6x=-2x(5x^{2}-2x - 3)\)
Factor \(5x^{2}-2x - 3\): \(5x^{2}-5x + 3x - 3 = 5x(x - 1)+3(x - 1)=(x - 1)(5x + 3)\)
So denominator is \(-2x(x - 1)(5x + 3)\)

Step2: Cancel common factor \((x - 1)\)

Now the limit becomes:
\(\lim_{x
ightarrow1}\frac{(x - 1)(x + 1)(5 - 2x)}{-2x(x - 1)(5x + 3)}\)
Cancel \((x - 1)\) (since \(x
ightarrow1\), \(x
eq1\) so we can cancel):
\(\lim_{x
ightarrow1}\frac{(x + 1)(5 - 2x)}{-2x(5x + 3)}\)

Step3: Substitute \(x = 1\)

Substitute \(x = 1\) into the simplified expression:
\(\frac{(1 + 1)(5 - 2\times1)}{-2\times1\times(5\times1 + 3)}=\frac{2\times3}{-2\times8}=\frac{6}{-16}=-\frac{3}{8}\)

Answer:

\(-\frac{3}{8}\)