Q.Evaluate $\lim_{x \to 1} \dfrac{x^7 - 2x^5 + 1}{x^3 - 3x^2 + 2}$.

Concept: Direct substitution for limits of rational functions; factorization when indeterminate.

First check direct substitution at $x = 1$:

• Numerator: $1^7 - 2(1)^5 + 1 = 1 - 2 + 1 = 0$
• Denominator: $1^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0$

The $\frac{0}{0}$ form means $(x-1)$ is a common factor. Factor it out from both:

Numerator: $x^7 - 2x^5 + 1 = (x-1)(x^6 + x^5 - x^4 - x^3 - x^2 - x - 1)$

Denominator: $x^3 - 3x^2 + 2 = (x-1)(x^2 - 2x - 2)$ …

Direct substitution gives $\frac{0}{0}$, so $(x-1)$ divides both polynomials.

Denominator: $x^3 - 3x^2 + 2 = (x-1)(x^2 - 2x - 2)$, and $x^2-2x-2$ at $x=1$ equals $-3\neq 0$, so $(x-1)$ is a simple factor.

Numerator: dividing $x^7 - 2x^5 + 1$ by $(x-1)$ gives

and the quotient at $x=1$ equals $1+1-1-1-1-1-1 = -3\neq 0$, so $(x-1)$ is again a simple factor. …