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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?

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$$\lim_{x\to \infty} \frac {x^x-4}{2^x-x^2}=\lim_{x\to \infty} \frac {x^x(1-\frac4{x^x})}{2^x(1-\frac{x^2}{2^x})}=\lim_{x\to \infty}\Big(\frac x2\Big)^x\cdot\Bigg[\frac{1-\frac4{x^x}}{1-\frac{x^2}{2^x}}\Bigg]\to\infty$$The latter tends to $1$, while the former tends to $\infty$.

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  • $\begingroup$ How to evaluate $\lim_{x\to \infty} \frac {1-\frac {4}{x^x}}{1-\frac {x^2}{2^x}}$ $\endgroup$ – Maggie Jan 9 '19 at 13:06
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    $\begingroup$ Both $\frac{4}{x^x}$ and $\frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $\frac{1}{1} = 1$. $\endgroup$ – KM101 Jan 9 '19 at 13:07
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We can say that $4\ll x^x$ and $x^2\ll2^x$ as $x$ grows large so

$$\lim_{x\to \infty} \frac {x^x-4}{2^x-x^2} \sim \lim_{x\to \infty}\frac{x^x}{2^x} = \lim_{x\to \infty} (\frac{x}{2})^x = \infty$$

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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 $\frac{1}{0^+} = +\infty$.

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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.

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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$${x^x-4\over 2^x-x^2}>{1\over 2}\left({x\over 2}\right)^x\to \infty$$

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