Why $\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}\neq 0$

Question

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?

Answer 1

The rule about powers in the numerator vs. powers in the denominator is only relevant when $x\to \pm \infty$.

For example, $$\lim_{x\to 3}\frac{x^2}{x^4}=\frac{3^2}{3^4}=\frac{1}{9}$$

Answer 2

That thing about the power of the polynomial works for limits at $\pm\infty$. That's not the case here.

Here, we have$$\lim_{x\to-1}\frac{x^3-2x-1}{x^4+2x+1}=\lim_{x\to-1}\frac{x^2-x-1}{x^3-x^2+x+1}=-\frac12.$$

Answer 3

$$ \left( x^{4} + 2 x + 1 \right) $$

$$ \left( x^{3} - 2 x - 1 \right) $$

$$ \left( x^{4} + 2 x + 1 \right) = \left( x^{3} - 2 x - 1 \right) \cdot \color{magenta}{ \left( x \right) } + \left( 2 x^{2} + 3 x + 1 \right) $$ $$ \left( x^{3} - 2 x - 1 \right) = \left( 2 x^{2} + 3 x + 1 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 3 }{ 4 } \right) } + \left( \frac{ - x - 1 }{ 4 } \right) $$ $$ \left( 2 x^{2} + 3 x + 1 \right) = \left( \frac{ - x - 1 }{ 4 } \right) \cdot \color{magenta}{ \left( - 8 x - 4 \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x \right) } \Longrightarrow \Longrightarrow \frac{ \left( x \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 2 x - 3 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x^{2} - 3 x + 4 }{ 4 } \right) }{ \left( \frac{ 2 x - 3 }{ 4 } \right) } $$ $$ \color{magenta}{ \left( - 8 x - 4 \right) } \Longrightarrow \Longrightarrow \frac{ \left( - 4 x^{3} + 4 x^{2} - 4 x - 4 \right) }{ \left( - 4 x^{2} + 4 x + 4 \right) } $$ $$ \left( x^{3} - x^{2} + x + 1 \right) \left( 2 x - 3 \right) - \left( x^{2} - x - 1 \right) \left( 2 x^{2} - 3 x + 4 \right) = \left( 1 \right) $$ $$ \left( x^{4} + 2 x + 1 \right) = \left( x^{3} - x^{2} + x + 1 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 0 \right) $$ $$ \left( x^{3} - 2 x - 1 \right) = \left( x^{2} - x - 1 \right) \cdot \color{magenta}{ \left( x + 1 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x + 1 \right) } $$ $$ \left( x^{4} + 2 x + 1 \right) \left( 2 x - 3 \right) - \left( x^{3} - 2 x - 1 \right) \left( 2 x^{2} - 3 x + 4 \right) = \left( x + 1 \right) $$

Answer 4

If you change: $x+1=t$, then: $$\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}=\lim_{t\to 0}\frac{(t-1)^3-2(t-1)-1}{(t-1)^4+2(t-1)+1}=\lim_{t\to 0}\frac{t^3-3t^2+t}{t^4-4t^3+6t^2-2t}=\\ \lim_{t\to 0}\frac{t^2-3t+1}{t^3-4t^2+6t-2}=-\frac12.$$