How can you prove that $\lim_{x \to \infty} \frac{x^{100}}{1.01^x} = 0$ using definition of limit?

Question

How can you prove that $\lim_{x \to \infty} \frac{x^{100}}{1.01^x} = 0$ Using the definition of a limit?

I have been dabbling with this problem, and I am struggling to find an answer. There seem to be limited online resources on this type of problem. The common proof is to say that exponentials grow faster than polynomials, but how would you solve it using the definition?

By defintion:
If $\lim_{x \to \infty} f(x) = A$. Then for every possible $\epsilon > 0$ there must exist a $ k > 0$ such that $| f(x) - A | < \epsilon$ if $x > k$.

So I have to find some function for the value of $k$ based on $\epsilon$ to prove the limit. I have tried:

$\bigg| \frac{x^{100}}{1.01^x} - 0 \bigg | < \epsilon$

$ \frac{x^{100}}{1.01^x} < \epsilon$

$ \log {\frac{x^{100}}{1.01^x}} < \log\epsilon$

$ 100*\log{x} - x*\log{1.01} < \log\epsilon$

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Now this is the part where I get stuck. I have tried to put in values for $x$, like $x=e$ the way blackpenredpen does: https://youtu.be/4pRMej3DnEM?t=512 and the continuing until I get x on the left side. This however seems to lead nowhere.

How would one solve this using the formal defintion?

Answer 1

Dealing with the function directly could be hard. Here is a way you can approach this. Ideally, you want

$$ \frac{x^{100}}{1.01^x}\leq \frac{1}{x}$$

Because if that is the case, then you can show that for $x\geq 1/\epsilon$

$$ \left|\frac{x^{100}}{1.01^x}\right|\leq \frac{1}{x} \leq \epsilon$$

Now, when is it the case that $ \frac{x^{100}}{1.01^x}\leq \frac{1}{x}$? Can you verify that this holds for sufficiently large $x$, say $x\geq 10^{10}$? If so, then for $k\geq \max(10^{10}, \frac{1}{\epsilon})$, we have

$$ \left|\frac{x^{100}}{1.01^x}\right|\leq \frac{1}{x} \leq \epsilon$$

And you're done.

Answer 2

These all come down to the fact that $\dfrac{\ln(x)}{x} \to 0 $ as $x \to \infty$.

More precisely, for any $c > 0$, there is an $x(c)$ such that $\dfrac{\ln(x)}{x} \lt c $ for $x > x(c) $.

If $f(x) =\dfrac{x^a}{b^x} $ where $a > 0, b>1$ then

$\begin{array}\\ g(x) &=\ln(f(x))\\ &=a\ln(x)-x\ln(b)\\ &=ax(\dfrac{\ln(x)}{x}-\dfrac{\ln(b)}{a})\\ \end{array} $

Once $x > x(\dfrac{\ln(b)}{2a}) $, $g(x) =ax(\dfrac{\ln(x)}{x}-\dfrac{\ln(b)}{a}) \lt ax(\dfrac{\ln(b)}{2a}-\dfrac{\ln(b)}{a}) = -x\dfrac{\ln(b)}{2} $ so $g(x) \to -\infty$ as $x \to \infty$ so $f(x) =e^{g(x)} \to 0 $.