Ytukyg

The textbook question is to solve this: $lim_{x to infty}=frac{ln(x)}{ln(x+sin(x))}$

I divide by $ln(x)$ and find :
$$lim_{x to infty}=frac{1}{1+frac{ln(sin(x))}{ln(x)}}$$

The answer is $1$ but i dont see how $lim_{x to infty}=1+frac{ln(sin(x))}{ln(x)}= 1$ because I thought $ln(sin(x)$ was undefined?

Maybe a possible simple way (without using things like Hospital rule):

$$
frac{ln{(x)}}{ln{(x+sin{(x)})}}=frac{ln{(x)}}{ln{(x(1+frac{sin{(x)}}{x}))}}=frac{1}{1+frac{ln{(1+frac{sin{(x)}}{x})}}{ln{x}}}
$$

Now what is the limit of
$$
lim_{xrightarrowinfty}frac{ln{(1+frac{sin{(x)}}{x})}}{ln{x}} = ?
$$

Then you can deduce the limit of:

$$
lim_{xrightarrowinfty}frac{1}{1+frac{ln{(1+frac{sin{(x)}}{x})}}{ln{x}}}
$$

This is important to realize that you can write
$$
lim_{xrightarrowinfty}frac{1}{1+frac{ln{(1+frac{sin{(x)}}{x})}}{ln{x}}}=frac{1}{1+lim_{xrightarrowinfty}frac{ln{(1+frac{sin{(x)}}{x})}}{ln{x}}}
$$
only because $yrightarrow frac{1}{1+y}$ is continuous (around the limit value).

You appear to have wrongly assumed that $$ln(x+sin(x))=ln(x)+ln(sin(x)). $$
An easy way to evaluate this limit is to use the known bounds on the sine function.

The answer that I gave (below) originally seemed reasonable. However, per back and forth with zhw. in the comments, I'm forced to conclude that his counterexample is valid. My original answer is therefore left as a flawed idea.

An alternative approach to Picaud Vincent's answer is to assume that if $dfrac{f(x)}{g(x)}rightarrow 1;$ as $xrightarrowinfty,;$ then $dfrac{text{ln}[f(x)]}{text{ln}[g(x)]}rightarrow 1.$

Let $f(x)=x;$ and $g(x)=x+text{sin}(x).;$ Since $text{sin}(x);$ is a bounded function, $;dfrac{f(x)}{g(x)}rightarrow 1;$ as $xrightarrowinfty.$