Types of Infinity — Question 2

Question 2 - Solution

We compare the growth rates of the polynomial $x^n$ and the exponential function $e^x$.

Both $x^n$ and $e^x$ are differentiable for all real $x$. Apply L’Hôpital’s Rule repeatedly to the limit $$\lim_{x\to\infty}\frac{x^n}{e^x}.$$

Differentiate the numerator and denominator once: $$\frac{d}{dx}(x^n)=n x^{\,n-1}, \qquad \frac{d}{dx}(e^x)=e^x.$$

Thus, $$\lim_{x\to\infty}\frac{x^n}{e^x} = \lim_{x\to\infty}\frac{n x^{n-1}}{e^x}.$$

Apply L’Hôpital’s Rule again. After $n$ applications, the numerator becomes a constant: $$\lim_{x\to\infty}\frac{n!}{e^x}.$$

Since $e^x$ grows without bound as $x\to\infty$, $$\lim_{x\to\infty}\frac{n!}{e^x}=0.$$

Therefore, $$\lim_{x\to\infty}\frac{x^n}{e^x}=0.$$

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