Why $\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}\neq 0$ According to solution, $\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}=\frac{1}{2}$. Why it is so, when power of polynomial in denominator is greater than in numerator?
 A: The rule about powers in the numerator vs. powers in the denominator is only relevant when $x\to \pm \infty$.
For example, $$\lim_{x\to 3}\frac{x^2}{x^4}=\frac{3^2}{3^4}=\frac{1}{9}$$
A: That thing about the power of the polynomial works for limits at $\pm\infty$. That's not the case here.
Here, we have$$\lim_{x\to-1}\frac{x^3-2x-1}{x^4+2x+1}=\lim_{x\to-1}\frac{x^2-x-1}{x^3-x^2+x+1}=-\frac12.$$
A: $$  \left(   x^{4}  + 2 x  + 1 \right)  $$ 
$$  \left(   x^{3}  - 2 x  - 1 \right)  $$ 
$$  \left(   x^{4}  + 2 x  + 1 \right)  =  \left(   x^{3}  - 2 x  - 1 \right)  \cdot \color{magenta}{  \left(   x  \right) } +  \left(  2 x^{2}  + 3 x  + 1 \right)  $$ 
 $$  \left(   x^{3}  - 2 x  - 1 \right)  =  \left(  2 x^{2}  + 3 x  + 1 \right)  \cdot \color{magenta}{  \left(   \frac{ 2 x  - 3 }{ 4 }  \right) } +  \left(   \frac{  -  x  - 1 }{ 4 }  \right)  $$ 
 $$  \left(  2 x^{2}  + 3 x  + 1 \right)  =  \left(   \frac{  -  x  - 1 }{ 4 }  \right)  \cdot \color{magenta}{  \left(   - 8 x  - 4 \right) } +  \left( 0 \right)  $$ 
 $$ \frac{ 0}{1} $$ 
 $$ \frac{ 1}{0} $$ 
 $$ \color{magenta}{  \left(   x  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   x  \right) }{ \left( 1  \right) } $$ 
 $$ \color{magenta}{  \left(   \frac{ 2 x  - 3 }{ 4 }  \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   \frac{ 2 x^{2}  - 3 x  + 4 }{ 4 }  \right) }{ \left(   \frac{ 2 x  - 3 }{ 4 }  \right) } $$ 
 $$ \color{magenta}{  \left(   - 8 x  - 4 \right) }  \Longrightarrow  \Longrightarrow  \frac{  \left(   - 4 x^{3}  + 4 x^{2}  - 4 x  - 4 \right) }{ \left(   - 4 x^{2}  + 4 x  + 4 \right) } $$ 
 $$  \left(   x^{3}  -  x^{2}  +  x  + 1 \right)  \left(  2 x  - 3 \right)  -  \left(   x^{2}  -  x  - 1 \right)  \left(  2 x^{2}  - 3 x  + 4 \right)  =  \left( 1  \right)  $$ 
 $$  \left(   x^{4}  + 2 x  + 1 \right)  =  \left(   x^{3}  -  x^{2}  +  x  + 1 \right)  \cdot \color{magenta}{  \left(   x  + 1 \right) } +  \left( 0 \right)  $$ 
 $$  \left(   x^{3}  - 2 x  - 1 \right)  =  \left(   x^{2}  -  x  - 1 \right)  \cdot \color{magenta}{  \left(   x  + 1 \right) } +  \left( 0 \right)  $$ 
 $$  \mbox{GCD} =   \color{magenta}{  \left(   x  + 1 \right) }   $$ 
 $$  \left(   x^{4}  + 2 x  + 1 \right)  \left(  2 x  - 3 \right)  -  \left(   x^{3}  - 2 x  - 1 \right)  \left(  2 x^{2}  - 3 x  + 4 \right)  =  \left(   x  + 1 \right)  $$ 
A: If you change: $x+1=t$, then:
$$\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}=\lim_{t\to 0}\frac{(t-1)^3-2(t-1)-1}{(t-1)^4+2(t-1)+1}=\lim_{t\to 0}\frac{t^3-3t^2+t}{t^4-4t^3+6t^2-2t}=\\
\lim_{t\to 0}\frac{t^2-3t+1}{t^3-4t^2+6t-2}=-\frac12.$$
