How can I approximate these two functions?

Question

I have a doubt: I have to compute a limit with Taylor series. In particular, my doubt is related to this composition of functions. $$\text{sin}(x + \text{log}(x+5))$$ [as $x\to 0$ from the right]. How can I approximate this sine, and this log?

As far as I know, I can't simply put Taylor expansions here, because: sine doesn't tend to zero, and log doesn't tend to 1. So, I can't use them here.

Is there a method I can use whenever I have "log(number not equal to 1 + x)" to approximate this stuff?

I'm confused.

Answer 1

In my humble opinion, when composing Taylor series, starting from the inside to the outside uses to make life easier.

Using your example $$\log(x+5)=\log (5)+\frac{x}{5}-\frac{x^2}{50}+O\left(x^3\right)$$ $$x+\log(x+5)=\log (5)+\frac{6 x}{5}-\frac{x^2}{50}+O\left(x^3\right)$$ $$\sin (x+\log (x+5))=\sin (\log (5))+\frac{6}{5} \cos (\log (5))x-\frac{36 \sin (\log(5))+\cos (\log (5))}{50} x^2 +O\left(x^3\right)$$

Trying for $x=0.1$, this would give $$\sin \left(\frac{1}{10}+\log \left(\frac{51}{10}\right)\right)\sim \frac{4964 \sin (\log (5))+599 \cos (\log (5))}{5000}=0.987431$$ while the exact value is $0.987474$.

Answer 2

$\sin{(x+\log(x+5))}= \sin{(x)}\cos{\log{(x+5)}}+\cos{x}\sin{(\log{(x+5)})}$ by the trig identity.

It's limit as $x\to 0$ is $\sin{(\log(x+5))}$

Assuming going forward that $log$ has base $e$...

Taylors' might be applicable if you start with the reciprocal of the log's argument.

$\ln{(x+5)}=-\ln{(\frac{1}{x+5})}\approx-\ln{(1-4/5)}\approx-1\cdot(-4/5)$

since $\ln(1-x)\approx -x$

$\sin{(4/5)} \approx 4/5-\frac{(4/5)^3}{3!}$