Question

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

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

Answer: (B)

$\frac{1}{3}$

$\begin{array}{c}\Rightarrow {lim}_{x\to -1}\frac{x^3-2x-1}{x^5-2x-1}\\ \Rightarrow {lim}_{x\to -1}\frac{\left(x+1\right)\left(x^2-x-1\right)}{\left(x+1\right)\left(x^4-x^3+x^2-x-1\right)}\\ \Rightarrow \frac{x^2-x-1}{\left(x^4-x^3+x^2-x-1\right)}\\ \Rightarrow \frac{2-1}{1+1+1+1-1}=\frac{1}{3}\end{array}$