I have 2 questions that I'm unsure how to answer, they are both related to same question and function.
I am given a function:
$$f(x)=\ln(\ln(1+\sin^{2}x)+1)+\cos (x) e^{\sin x}-\frac{\ln(x+1)}{e^{x^{2}+x+1}}$$
Q1. Write down the limit expression for the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$.
Q2. For the given function $f$ determine if the claim that $g(x)=x+1$, is an approximation of $f(x)$ of order $1$ near $0$ is true. Justify your answer.
Any help on this would be most grateful, as I have no idea what it's getting at or how to answer the above.
Thank you so much for your help in advance.
We have that $$f(x)=\ln(\ln(1+sin^{2}x)+1)+cos(x)e^{sinx}-\frac{ln(x+1)}{e^{x^{2}+x+1}}$$ and so $f(x)\approx f(0)+xf'(0).$ Since $$f'(0) = \frac{e-1}{e}$$ and $$f(0) = 1$$ we get that $$f(x)\approx 1 + x\left(\frac{e-1}{e}\right) = 1+x-x/e.$$
So for question $1$ if $g$ is an approximation of $f$ then it must be that $$f(x)\approx f(0) + xf'(0)=1+x $$ and so it must be the case that $$\lim_{x\to 0}\frac{f(x)}{1+x}=1.$$
For the second question, you can see that since $f'(0)\neq 1$, $g(x)$ is not an approximation of $f$ near $0.$