Graph: $$ \lim _{n \rightarrow \infty} \frac{{|x+1 \mid}^{n}+x^{2}}{|x|+x^{2n}} $$
Here I tried first to resolve the limit, dividing by ${x}^{2n}$ in both numerator and plugging in $\infty$ I obtain the limit to be zero for all values of $x$. But when it tried to graph it using a graphing calculator I obtained the following graph (I put n=10000),
Can anyone please point out my mistake?
Case by case:
If $x>\varphi = \frac{1+\sqrt{5}}{2}$, then $x+1<x^2$ and we can divide by $x^{2n}$ to se that:
$$ \lim_n \frac{(x+1)^n + x^2}{x+(x^2)^n}=\lim_n \frac{(\frac{x+1}{x^2})^n + x^{2-2n}}{x^{1-2n}+1}=\frac{0+0}{0+1}=0 $$
If $0<x<\varphi$, then $x+1>x^2$ , thus
$$ \lim_n \frac{(x+1)^n + x^2}{x+(x^2)^n}=\lim_n \frac{(\frac{x+1}{x^2})^n + x^{2-2n}}{x^{1-2n}+1}=\frac{\infty+0}{0+1}=\infty $$
If $-1<x<0$, then $x^2<1$ and $0<x+1<1$, and the limit turns out to be: $$ \lim_n \frac{(x+1)^n + x^2}{-x+(x^2)^n}=\frac{x^2}{-x}=-x $$
If $-2<x<-1$, then $|x+1|<1$ and $x^2>1$, resulting in
$$ \lim_n \frac{|x+1|^n + x^2}{-x+(x^2)^n}=\frac{0+x^2}{-x+\infty}=0 $$
And finally, for $x<-2$ we can divide by $x^2$, and taking into account that $\frac{|x+1|}{x^2}<1$ we get that
$$ \lim_n \frac{|x+1|^n + x^2}{-x+(x^2)^n}=\lim_n \frac{(\frac{-(x+1)}{x^2})^n + x^{2-2n}}{x^{1-2n}+1}=\frac{0+0}{0+1} = 0 $$
Maybe we could have considered the last two together, but I think that separating them helps conceptually.