Consider the function $3^x(1-2^{-\text{frac}(x\sqrt2)})$, where $\text{frac}()$ denotes the fractional part function. Now $\sqrt 2$ being irrational by kronecker, this is uniformly dense in $(0,1)$, which means that $\text{frac}(x\sqrt2)$ can get very close to zero.
I want to prove that for sufficiently large positive integers $x$, this function is bigger than any specified lower bound. To be very specific, for any positive real $L$, prove that there is a positive integer $N$, such that the function evaluated at any positive integer greater than $N$, gives a value bigger than $L$. Now the above thing causes an issue. The above clearly seems to be true by Wolfram and it is intuitively clear this should be true,in a sense that even if the fractional part approaches zero, the exponential should dominate it and the function shouldn't experience a "sharp drop" over the positive integers. It is fairly easy to prove that it is unbounded again by kronecker but i have no idea how to prove the above.
What you need is a lower bound for the approximation of $\sqrt2$ by fractions.
Lemma. Suppose $x,y$ are positive integers. Then $$\left|\sqrt 2x-y\right|>\frac1{5x}$$ Proof. There are two cases.
Then $$\sqrt 2x-y<\sqrt2x-3x<-x\le-1$$ So, $|\sqrt2x-y|>1>\frac1{5x}$.
Since $y$ is arbitrary, the lemma implies $\text{frac}(x\sqrt2)>\frac1{5x}$ for any positive integer $x$.
$$\begin{aligned} 3^x(1-2^{-\text{frac}(x\sqrt2)}) &\ge3^x(1-2^{-\frac1{5x}})\\ &=3^x(1-e^{-\frac{\ln 2}{5x}})\\ &>3^x(1-(1+\frac12(-\frac{\ln 2}{5x}))\\ &=\frac{\ln2}{10}\frac{3^x}{x}\\ \end{aligned} $$ which goes to $\infty$ when $x$ goes to $\infty$.
The last inequality above uses the following inequality: $e^z<1+\frac z2$ when $z\in(-\frac12,0)$.