# 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?

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

• How to evaluate $\lim_{x\to \infty} \frac {1-\frac {4}{x^x}}{1-\frac {x^2}{2^x}}$ – Maggie Jan 9 '19 at 13:06
• Both $\frac{4}{x^x}$ and $\frac{x^2}{2^x}$ tend to $0$ for large $x$. So it becomes $\frac{1}{1} = 1$. – KM101 Jan 9 '19 at 13:07

We can say that $$4\ll x^x$$ and $$x^2\ll2^x$$ as $$x$$ grows large so

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

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

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.

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