Limit with logarithms

Question

Can you please help to calculate the following limit $$\lim_{x \to \infty} \left(\frac{\ln(x\ln(\gamma) )}{\ln(x\ln(\gamma)+\ln(\frac{\gamma-1}{\gamma}))}\right)^{1/2},$$ where $\gamma >10$ is a constant. It's going to be $1$, however I am not sure to prove it formally. What is more I would like to ask if I can write that
$$\lim_{x \to \infty} \left(\frac{\ln(x\ln(\gamma) )}{\ln(x\ln(\gamma)+\ln(\frac{\gamma-1}{\gamma}))}\right)^{1/2} = \lim_{x \to \infty}\left(\frac{\ln(x\ln(\gamma) )}{\ln(x\ln(\gamma)+\ln(\frac{\gamma-1}{\gamma}))}\right)^{1/2} $$ and then use Hospital Rule?

Answer 1

Put $\ln\gamma=a$; $\ln\frac{\gamma-1}{\gamma}=b$ so one has $$lim_{x\to \infty}\left(\frac{\ln(ax)}{\ln(ax+b)}\right)^{\frac 12}=lim_{x\to \infty}\left(\frac{\frac {a}{ax}}{\frac{a}{ax+b}}\right)^{\frac 12}=lim_{x\to \infty}\left(\frac{ax+b}{ax}\right)^{\frac 12}=1$$

Answer 2

Yes you can since $x\mapsto \sqrt x$ is a continuous function.

Then there is hardly anything left:

$\dfrac{\ln (x\ln \gamma)}{\ln (x\ln \gamma +\ln \dfrac{\gamma-1}{\gamma})}=\frac{\infty}{\infty}$..

Applying L-Hospital Rule reduces to :

$\dfrac{\ln{\frac{\gamma-1}{\gamma}} +x\ln \gamma }{x\ln \gamma}$

Applying L-Hospital Rule reduces to : $1$

Answer 3

One option: write $t=x\ln\gamma$ for convenience, so that $t\to \infty$; and set $\alpha=\ln \frac{\gamma-1}{\gamma}$ for conciseness as well. Then, you have, using Taylor expansions (first order; detailing each step) $$\begin{align} \left(\frac{\ln t}{\ln(t+\alpha)}\right)^{1/2}&= \left(\frac{\ln t}{\ln t+\ln(1+\frac{\alpha}{t})}\right)^{1/2} = \left(\frac{\ln t}{\ln t+\frac{\alpha}{t} + o\left(\frac{1}{t}\right)}\right)^{1/2}\\ &= \left(\frac{1}{1+\frac{\alpha}{t\ln t} + o\left(\frac{1}{t\ln t}\right)}\right)^{1/2}\\ &= \left(1-\frac{\alpha}{t\ln t} + o\left(\frac{1}{t\ln t}\right)\right)^{1/2}\\ &= 1-\frac{\alpha}{2t\ln t} + o\left(\frac{1}{t\ln t}\right)\\ &\xrightarrow[t\to\infty]{}1. \end{align}$$ (We used that $\ln(1+u) = u+o(u)$, and $(1+u)^a = 1+au+o(u)$, when $u\to0$.)

Note that this actually gives a bit more than the limit: it also tells you about the second term of the asymptotic behavior.

Answer 4

Using the properties of logarithms, and in particular $\ln(ab) = \ln a + \ln b$, we can rewrite the limit as follows: $$\lim_{x \to +\infty} \left(\frac{\ln x + \ln\ln(\gamma)}{\ln x + \ln\left(\ln(\gamma)+\frac1x\ln(\frac{\gamma-1}{\gamma})\right)}\right)^{1/2}.$$

Now factor out $\ln x$ from both the numerator and the denominator and you'll be left with something of the form $$\lim_{x \to +\infty} \sqrt{\frac{1 + o(x)}{1 + o(x)}}$$ which evaluates to $1$ since $x \mapsto \sqrt x$ is continuous.