Approximation in the neighborhood of infinity to a sum of trigonometric and exponential function

Question

The function to approximate in the neighborhood of $\infty$ is $f(x)=e^{\frac{1}{x^2}}-\cos(\frac{1}{x^2})-\sin(\frac{1}{x^2})$. I know that if $f \sim \phi$ and $g \sim \psi$ then $f+g \sim \phi + \psi$ only when $\phi$ and $\psi$ have the same sign in the neighborhood of $\infty$. But I was not sure about whether $f-g \sim \phi + \psi$ is true. So I let $u=\frac{1}{x}$ so $f(u)=e^{u^2}-\cos(u^2)-\sin(u^2)$. I find that at $u=0$, $f=f'=f^{\prime\prime}=f^{\prime\prime\prime}=0$,and at $u=0$ $f^{(4)}=24$. so $f(x) \sim \frac{24}{x^4}$ which is not the result given by Wolfram Taylor series calculator. I let then $u=\frac{1}{x^2}$, I get now $f(x) \sim \frac{2}{x^4}$. Also the wrong result which should be $f(x) \sim \frac{1}{x^4}$. What did I do wrong? Also if $f \sim \phi$ and $g \sim \psi$ then is it true that $f-g \sim \phi - \psi$ only when $\phi$ and $\psi$ have the same sign in the neighborhood of $\infty$. Can I apply it in this case if it is true?

Answer 1

You forgot the factor of $n!$ in the Taylor expansion: $f(x)\sim\frac{f^{(4)}(0)}{4!}u^4=\frac1{x^4}$.

As to your question about the signs: The restriction to the same sign is substantial and you can’t get around it with this trick of relabeling $g$ and $\psi$ to include the sign. For $f=\frac1x$, $\phi=\frac1x+\frac1{x^2}$, and $g=\psi=-\frac1x$, we have $0=f+g\not\sim\phi+\psi=\frac1{x^2}$, and applying your trick would lead to $g=\psi=\frac1x$, again with $0=f-g\not\sim\phi-\psi=\frac1{x^2}$, despite $\phi$ and $\psi$ now having the same sign.