determine which dominates $e^x$ or $3^{\sqrt{x}}$

Question

Is there a way to prove that $e^x$ dominates $3^{\sqrt{x}}$ as x goes to infinity using Calculus II level math. I was hoping to prove it using L'Hopital's rule. As I applied L'Hopital's rule it seems apparent that both functions will derivate infinitely many times. However by going through two iterations I arrived at $$\lim _{x\to \infty }\left(\frac{e^x}{3^{\sqrt{x}}}\right) = \ln\left(3\right)\cdot \lim _{x\to \infty }\left(\frac{e^x2\sqrt{x}}{3^{x^{\frac{1}{2}}}}\right)+4\:\ln\left(3\right) \lim _{x\to \infty }\left(\frac{\left(e^x\cdot x\cdot \sqrt{x}\right)}{3^{x^{\frac{1}{2}}}}\right)$$ (hopefully there weren't any errors when I took the derivatives) It seems to me that the numerator will become infinitely more complex as we continue to take derivatives and the denominator will remain the same (by "kicking" terms up to the numerator). To me this seems to suggest that $e^x$ dominates $3^{\sqrt{x}}$. How would I show that reasoning mathematically. More over does my assumption/intuition that a function whose derivatives balloon and become infinitely more complex as you take L'Hopital's rule over and over again prove the limit goes to infinity?

I welcome any suggestions on how to clarify my question in a more mathematical way. As well as a proof of which function dominates using other means than L'Hopital's rule.

Answer 1

$$\begin{align*} \lim_{x\to \infty} \frac{e^x}{3^{\sqrt x}}&=\lim_{x\to \infty} \frac{(e^{\sqrt x})^{\sqrt x}}{3^{\sqrt x}}\\ &=\lim_{x\to \infty} \left( \frac{e^{\sqrt x}}{3}\right)^{\sqrt x}\\ &=\left(\to \infty\right)^{\to \infty} = \infty \end{align*}$$

Answer 2

Applying the suggestions above (while avoiding hand grenades ;) I've come up with: $$\lim _{x\to \infty }\left(\frac{\left(e^x\right)}{3^{\sqrt{x}}}\right) = \lim _{x\to \infty }\frac{e^x}{e^{x^{\left(\frac{1}{2}\right)}\cdot \:ln\left(3\right)}} = \lim _{x\to \infty }e^{x-x^{\left(\frac{1}{2}\right)}\cdot \ln\left(3\right)}$$ Then applying the limit chain rule to consider $\lim _{x\to \infty }{x-\ln\left(3\right)\cdot \:\:\:\:x^{\left(\frac{1}{2}\right)}} $ by factoring out an x we get $\infty \cdot \:1$. We can then reconsider $\lim _{x\to \infty }e^{x-x^{\left(\frac{1}{2}\right)}\cdot \:ln\left(3\right)}$ with $e^\infty$ is $\infty$ therefore $e^x$ dominates

I welcome any suggestions on how to beautify the above.

Answer 3

Taking the natural logarithm of the ratio,

$$x-\log3\sqrt x=\sqrt x(\sqrt x-\log3)$$ which certainly tends to infinity.

Answer 4

For each $x \in \mathbb R^{+}$ we can write $\frac{e^x}{3^{\sqrt x}}=(\frac{e^{\sqrt x}}{3})^{\sqrt x}$

Now, for $x≤(\text{ln}3)^2=1.2069... ,$ $(\frac{e^{\sqrt x}}{3})≤1 \rightarrow (\frac{e^{\sqrt x}}{3})^{\sqrt x}≤1$

So, for $0<x<(\text{ln}3)^2, e^x$ is dominated by $3^{\sqrt x}$. At $x=0,(\text{ln}3)²$ they are equal. And for $x>(\text{ln}(3))²$ $e^{x}$ dominates $3^{\sqrt x}$.

Now, when $x<0$, $\sqrt x$ has no existence in real domain. But if it were be like $\sqrt {|x|}$ instead of $\sqrt x$, then surely $3^{\sqrt {|x|}}>3^{0}=e^{0}>e^{x}$ for $x<0$. So, in this case $3^{\sqrt {|x|}}$ would dominate $e^{x}$ in negative domain.