Why $\lim_{n\to \infty} \frac{(2^2)^{\sqrt n}}{(1+(10)^{-2^2})^n}$ has no limit?

Question

$\lim_{n\to \infty} \frac{(2^2)^{\sqrt n}}{(1+(10)^{-2^2})^n}$

I looked it up in the wolfram and my intuition tells me that this expression has no limit, but every step I did doesn't lead me to the solution.

Answer 1

Why do you come to that conclusion? You are looking at $$ \frac{4^\sqrt{n}}{(1+10^{-4})^n}$$

Now notice that $4>1$ and $1+10^{-4}>1$, so both denominator and numerator are growing, but as $\sqrt n$ grows much slower than $n$ the numerator will also grow slower than the denominator, so the whole thing goes to $0$.

To see this simply apply a logarithm (to get rid of the different bases):

$$\log \ldots = \log(4) \sqrt{n} - \log(1+10^{-4})n $$ Obviously this goes to $-\infty$ as $n\to\infty$.

Answer 2

Observe that for $n$ big enough we have

$$5<(1+10^{-4})^\sqrt{n}$$ so $$5^{\sqrt{n}}<(1+10^{-4})^n $$ and then $$\frac{4^{\sqrt n}}{(1+10^{-4})^n} <\frac{4^{\sqrt n}}{5^{\sqrt{n}}}=\left(\frac{4}{5}\right)^{\sqrt n} \to 0$$