Only the letter g) and I know it’s infinity - infinity but I don’t know how to modify it to be able to find the answer

You have to use some properties of logarithms. I'm assuming the logarithm is real-valued, in which case the following hold:

[tex]\ln a-\ln b=\ln\dfrac ab[/tex]
[tex]\ln a^n=n\ln a[/tex]

for real [tex]a,b>0[/tex] and all real [tex]n[/tex].

So we can write

[tex]\ln\sqrt{9x+2}-\ln\sqrt{4x+5}=\ln(9x+2)^{1/2}-\ln(4x+5)^{1/2}[/tex]
[tex]=\ln\dfrac{(9x+2)^{1/2}}{(4x+5)^{1/2}}[/tex]
[tex]=\ln\left(\dfrac{9x+2}{4x+5}\right)^{1/2}[/tex]
[tex]=\dfrac12\ln\dfrac{9x+2}{4x+5}[/tex]

In taking the limit, we're considering [tex]x[/tex] as it gets arbitrarily large. We know that for [tex]x>0[/tex], [tex]\ln x[/tex] is continuous. This means we can pass the limit through the logarithm:

[tex]\displaystyle\lim_{x\to\infty}\frac12\ln\dfrac{9x+2}{4x+5}=\frac12\ln\left(\lim_{x\to\infty}\frac{9x+2}{4x+5}\right)[/tex]

so now we're only concerned with the limit of a rational function. The leading terms in the numerator and denominator both have the same power, so we only need to consider their coefficients. In other words,

[tex]\dfrac{9x+2}{4x+5}\approx\dfrac{9x}{4x}=\dfrac94[/tex]

when [tex]x\neq0[/tex], and so the limit is the same as

[tex]\displaystyle\frac12\ln\left(\lim_{x\to\infty}\frac94\right)=\frac12\ln\frac94=\ln\left(\frac94\right)^{1/2}=\ln\frac32[/tex]