Repeated application of L'Hopital's rule to evaluate $\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$

Question

I came across an interesting limit while trying to draw the graph of the function $f(x)=\frac{3^{\ln x}}{x}$.

What is $$L=\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$$

Since $\ln(0^+)\to -\infty$ and $3^{-\infty}=0$ we have the limit in indeterminate form of $\frac{0}{0}$. So By L'Hopital's rule we get $$L=\lim_{ x \to 0^+}\frac{3^{\ln x}\ln 3 \times \frac{1}{x}}{1}=\ln 3\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$$ So how to find this limit?

Answer 1

$$ L = (\ln3)\cdot L $$

Dividing both sides by $L$ (if $L\ne0$) yields $1 = \ln 3,$ which is false.

So the only way the first line above can be true is if $L=0.$

Answer 2

Writing $x=e^{-t}$ and considering $t\to +\infty$ you get

$$\frac{3^{\ln x}}{x} = \left(\frac{3}{e}\right)^{-t}=\left(\frac{e}{3}\right)^{t}\stackrel{t\to +\infty}{\longrightarrow}0$$

Answer 3

$\dfrac{3^{\log x}}{e^{\log x}} = (3/e)^{\log x};$

Take the limit $x\rightarrow 0^+.$

Answer 4

ok, Here's another way: $$ \overbrace{\frac{3^{\ln x}} x = \frac{x^{\ln 3}} x}^\text{by the change-of-base identity} \!\!\!\!\!\!\!\!\! = x^{(\ln3)-1} \quad \text{and } (\ln3)-1>0. $$ And so on.