Find $\lim_{x\to\infty}(\ln(1+x^2))/(\lg(10+x^4))$ without using the L'Hopital's rule or Taylor's series

Question

This limit is proposed to be solved without using the L'Hopital's rule or Taylor series: $$ \lim_{x\to\infty}\frac{\ln(1+x^2)}{\lg(10+x^4)}, $$ where $\lg x=\log_{10}x$. I know how to calculate this limit using the L'Hopital's rule: $$ \lim_{x\to\infty}\frac{\ln(1+x^2)}{\lg(10+x^4)}= \lim_{x\to\infty}\frac{2x}{1+x^2}\frac{10+x^4}{4x^3}\ln 10 =\lim_{x\to\infty}\frac{\ln 10}{1+x^2}\frac{10+x^4}{2x^2} $$ $$ =\lim_{x\to\infty}\frac{x^4(1+\frac{10}{x^4})}{2x^4(1+\frac1{x^2})}\ln 10 =\frac{\ln 10}2, $$ but I have no idea how to calculate this limit without derivatives, etc.

Answer 1

$$\frac{\ln(1+x^2)}{\lg(10+x^4)}=\log (10)\frac{ \log \left(x^2+1\right)}{\log \left(x^4+10\right)}=\log (10)\frac{2\log(x)+\log \left(1+\frac{1}{x^2}\right)}{4\log(x)+\log \left(1+\frac{10}{x^4}\right) }$$

Answer 2

Hint: Show for $a$ constant, $$\lim_{y\to\infty}\frac{\ln(a+y)}{\ln y}=1.$$

Use that to show your limit is the same as $$\lim_{x\to\infty}\frac{\ln(x^2)}{\log_{10}(x^4)}$$ But that limit is constant.

Second Hint:$$\ln(a+y)=\ln(y)+\ln\left(1+\frac ay\right)$$

Answer 3

You could transform the logs, i.e change $\log_{10}$ to $\ln$ or $\ln$ to $\log_{10}$ using the $\log$ base change formula and then continue onwards. Alternatively, you could argue using the definitions of asymptotic growth (you can have a little read at Big O notation) that in such case the base of $\log$ doesn’t matter when $\log(\dots) \rightarrow \infty$ in order to simplify your limit i.e we have $$\lim_{x\to\infty} \frac{\log(1+x^2)}{\log(10+x^4)}$$ (Note that we don’t have to specify the base as to corroborate our point)