Evaluate $\lim\limits_{x\to \infty} \frac {x^x-4}{2^x-x^2}$ Evaluate $$\lim_{x\to \infty} \frac {x^x-4}{2^x-x^2}$$
I think it needs to use L'Hospital Rule.
So, I first calculate $\frac {d x^x}{dx}=
x^x(\ln x+1)$.
And then $$\lim_{x\to \infty} \frac {x^x-4}{2^x-x^2}=\lim_{x\to \infty} \frac {x^x(\ln x+1)}{2^x(\ln 2)-2x}$$
It seems that I need to use L'Hospital Rule again. But when I do it, the thing inside the limit becomes more complicated.
How should I do next? Or maybe my way is false? 
 A: $$\lim_{x\to \infty} \frac {x^x-4}{2^x-x^2}=\lim_{x\to \infty} \frac {x^x(1-\frac4{x^x})}{2^x(1-\frac{x^2}{2^x})}=\lim_{x\to \infty}\Big(\frac x2\Big)^x\cdot\Bigg[\frac{1-\frac4{x^x}}{1-\frac{x^2}{2^x}}\Bigg]\to\infty$$The latter tends to $1$, while the former tends to $\infty$.
A: We can say that $4\ll x^x$ and $x^2\ll2^x$ as $x$ grows large so
$$\lim_{x\to \infty} \frac {x^x-4}{2^x-x^2} \sim \lim_{x\to \infty}\frac{x^x}{2^x} = \lim_{x\to \infty} (\frac{x}{2})^x = \infty$$
A: You’re better off avoiding L’Hôpital when solving limits. Keep it simple and divide all the terms by $x^x$:
$$\lim_{x\to \infty} \frac {1-\frac{4}{x^x}}{\frac{2^x}{x^x}-\frac{x^2}{x^x}}$$
And now, you can tell that $\frac{4}{x^x}$, $\frac{2^x}{x^x}$, and $\frac{x^2}{x^x}$ all tend to $0$, and since $2^x > x^2$ as $x$ grows large, the denominator tends to $0^+$. Hence, the limit becomes $\frac{1}{0^+} = +\infty$.
A: Quick & dirty (though correct) method:
$4$ and $x^2$ are neglectible in front of the exponentials. Then the expression is asymptotic to
$$\left(\frac x2\right)^x$$ which grows unboundedly.
A: Hint 
for $x$ being sufficiently large ($x>5$) we have
 $$2^x-x^2<2^x$$and $$x^x-4>{x^x\over 2}$$therefore$${x^x-4\over 2^x-x^2}>{1\over 2}\left({x\over 2}\right)^x\to \infty$$
