Question

Solve the following question

Answer 1

$Given : \lim_{n \to 2} ( \frac{1}{x - 2} - \frac{2(2x - 3)}{x^3 - 3x^2 + 2x} )$

$= \ \textgreater \ \frac{1}{x - 2} - \frac{4x - 6}{x(x - 1)(x - 2)}$

$= \ \textgreater \ \frac{1 * x(x - 1)}{x(x - 2)(x - 1)} - \frac{4x - 6}{x(x - 2)(x - 1)}$

$= \ \textgreater \ \frac{x(x - 1) - (4x - 6)}{x(x - 2)(x - 1)}$

$= \ \textgreater \ \frac{x^2 - x - 4x + 6}{x(x - 2)(x - 1)}$

$= \ \textgreater \ \frac{x^2 - 5x + 6}{x(x - 2)(x - 1)}$

$= \ \textgreater \ \frac{(x - 2)(x - 3)}{x(x - 2)(x - 1)}$

$= \ \textgreater \ \frac{x - 3}{x(x - 2)}$

$\frac{x - 3}{x^2 - 2x}$

Now,

$= \ \textgreater \ \lim_{n \to 2} \frac{x - 3}{x^2 - x}$

$= \ \textgreater \ \frac{2 - 3}{(2)^2 - 2}$

$= \ \textgreater \ \frac{-1}{4 - 2}$

$= \ \textgreater \ \frac{-1}{2}$




Hope this helps!