How to solve non-fractional limits like $\lim_{x \to \infty} x^{(\ln5) \div (1+\ln x)}$? I have the limit $\lim_{x \to \infty} x^{(\ln5) \div (1+\ln x)}$. I am trying to figure out how to solve this, but I only know how to handle limits when they can be made into fractions. Is there some way I can do that here, or should I try something else?
This is a homework problem, and I don't need a complete answer, but I'd appreciate some advice.
 A: You can use the exponential $$\exp(\ln f(x))= f(x)$$and the fact that $$\lim_{x\rightarrow a} e^{f(x)}=e^{\lim\limits_{x\to a} f(x)}$$ then use natural log properties
A: Let $P = x^{\frac {\ln (5)} {1 + \ln (x)}}.$ Then $\ln P = \frac {\ln (5)\ln (x)} {1 + \ln (x)}.$  As $x \rightarrow \infty,$ $\ln P$ takes $\frac {\infty} {\infty}$ form. So by L'Hospital rule we have $$\lim\limits_{x \to \infty} \ln P = \lim\limits_{x \to \infty} \frac {\ln (5) \cdot \ln (x)} {1 + \ln (x)} = \ln (5) \cdot \lim\limits_{x \to \infty} \frac {\frac 1 x} {\frac 1 x} = \ln (5).$$ This shows that $$\lim\limits_{x \to \infty} P = e^{\ln(5)} = 5.$$
A: I think the standard method is to write it with $e$:
$\lim_{x\to\infty} x^{\frac{\ln(5)}{1+\ln(x)}}=\lim_{x\to\infty} e^{\ln(x)\ln(\frac{\ln(5)}{1+\ln(x)})}$
A: We asked to evaluate $\lim \limits_{x \rightarrow \infty} x^{\frac{\ln(5)}{1+\ln(x)}}$ which can also be written as $\lim \limits_{x \rightarrow \infty} e^{\ln(x){\frac{\ln(5)}{1+\ln(x)}}}$. Using a few other rules, you can get to the following expression
$$ e^{\ln(5) \cdot \lim \limits_{x \rightarrow \infty} \frac{\ln(x)}{1+\ln(x)} } $$
Now I will leave the details to you but as a hint, remind yourself of L'hospital's rule.
And finally, you can show:
$$ \lim \limits_{x \rightarrow \infty} x^{\frac{\ln(5)}{1+\ln(x)}} = 5 $$
