Question

Evaluate the following limits:

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

Answer 1

When x = 1, the expression $\frac{x^7-2x^5+1}{x^3-3x^2+2}$ assumes the form $\frac{0}{0}$. So, (x − 1) is a factor of numerator and denominator.

Using long division method, we get

$x^7-2x^5+1=\left(x-1\right)\left(x^6+x^5-x^4-x^3-x^2-x-1\right)$

and $x^3-3x^2+2=\left(x-1\right)\left(x^2-2x-2\right)$

$$\begin{gathered} \therefore \underset{x\to 1}{\lim }\frac{x^7-2x^5+1}{x^3-3x^2+2} \\ =\underset{x\to 1}{\lim }\frac{\left(x-1\right)\left(x^6+x^5-x^4-x^3-x^2-x-1\right)}{\left(x-1\right)\left(x^2-2x-2\right)} \\ =\underset{x\to 1}{\lim }\frac{x^6+x^5-x^4-x^3-x^2-x-1}{x^2-2x-2} \\ =\frac{1+1-1-1-1-1-1}{1-2-2} \\ =\frac{-3}{-3} \\ =1 \end{gathered}$$