How to calculate limit of logarithm with variable base and argument

Let's say I have to calculate limit of this form: $$\lim_{x\to\infty}\frac{\log (x^3+x^2+x+1)}{\log(x^2+x+1)}$$

Can I do the following operation:

$$ \lim_{x\to\infty}\frac{\log (x^3+x^2+x+1)}{\log(x^2+x+1)} = \lim_{x\to\infty}\frac{\log (x^3(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}))}{\log (x^2(1+\frac{1}{x}+\frac{1}{x^2}))} = \lim_{x\to\infty}\frac{\log x^3}{\log x^2} $$ Because for large $x$ smaller powers are not that important and then I do: $$ \lim_{x\to\infty}\frac{\log x^3}{\log x^2}=\lim_{x\to\infty}{\log_{x^2}{x^3}}=\frac{3}{2} $$ Is this the correct way to solve this type of limits if I can only use simple transformations (so no derivatives)?

Solution 1:

Your method works well but I think we can argue slightly more reigourously using some properties of $\log$

$$ \lim_{x\to\infty}\frac{\log (x^3+x^2+x+1)}{\log(x^2+x+1)} = \lim_{x\to\infty}\frac{\log (x^3(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}))}{\log (x^2(1+\frac{1}{x}+\frac{1}{x^2}))} = \lim_{x\to\infty}\frac{\log (x^3) + \log(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3})}{\log (x^2) + \log(1+\frac{1}{x}+\frac{1}{x^2})} $$

Taking the limits of the $\log$ terms on the right of the numerator and denominator gives $0$ so we are left with the expression you obtained. matheusbm98 also gave an alternative method to solve this last bit which I like as well

$$\lim_{x\to\infty}\frac{\log (x^3)}{\log (x^2)} = \lim_{x\to\infty}\frac{3\log x}{2\log x} = \frac{3}{2} $$