Question

${lim}_{x\to 1}\frac{x^7-2x^5+1}{x^3-3x^2+2}$

Answer 1

Putting x =1 in the numerator and denominator fo the expression $\frac{x^7-2x^5+1}{x^3-3x^2+2}$, it takes $\left(\frac{0}{0}\right)$ form. This means (x-1)is a factor of both numerator and denominator. So, for factorising, divide both numerator and denominator by (x-1)as follow.

So, $x^7-2x^5+1=(x-1)(x^6+x^5-x^4-x^3-x^2-x-1)$ by division algorithm.

Agian,

$\therefore {lim}_{x\to 1}\frac{x^7-2x^5+1}{x^3-3x^2+2}$ [It is of $\left(\frac{0}{0}\right)$ form]

$={lim}_{x\to 1}\frac{(x-1)(x^6+x^5-x^4-x^3-x^2-x-1)}{(x-1)(x^2-2x-2)}$

[It is of $\left(\frac{0}{0}\right)$ form]

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

$=\frac{(1{)}^6+(1{)}^5-(1{)}^4-(1{)}^3-(1{)}^2-1-1}{(1{)}^2-2(1)-2}$

$=\frac{-3}{-3}=1$