for $f(x)=x \log(e+\frac{1}{x}) (x>0)$ ,
answer this questions:
1.) $\alpha=\lim\limits_{x \to \infty} {\frac{f(x)}{x}}$, find $\alpha$
2.) from result $\alpha$ from number 1, $\beta=\lim\limits_{x \to 0^+} {(f(x)-\alpha(x))}$ and $t=\frac{1}{x}$, $x\to \infty$ and $t\to 0^+$ find $\beta$
3.) from result $\alpha$ and $\beta$ in number 1 and 2, find $\lim\limits_{x \to \infty} \frac{f(x)-(\alpha x+\beta)}{\frac{1}{x}}$
attempt:
1.)$\alpha=\lim\limits_{x \to \infty} {\frac{f(x)}{x}}$= $\frac{x \log(e+\frac{1}{x}) }{x}=\log(e+\frac{1}{x})$=$\log e$
2.) $\beta=\lim\limits_{x \to \infty} {(f(x)-\alpha(x))}$,
if $t=\frac{1}{x}$ , then $x\to\infty$ $t\to0$, change limit to $t$
$\beta=\lim\limits_{x \to \infty} {(f(x)-\alpha(x))}$=$\lim\limits_{x \to \infty} {x\log(e+\frac{1}{x})-x \log e}$
$=\lim\limits_{x \to 0} {\frac{1}{t}\ln(e+t)-\frac{1}{t}\ln e}$
=$$\beta=\lim\limits_{t \to 0}{\frac{\ln(\frac{e+t}{e})}{t}}=0$$
3.) from $\alpha $ and $\beta$ in number 1 and 2
$\lim\limits_{x \to \infty} \frac{f(x)-(\alpha x+\beta)}{\frac{1}{x}}$ = $\lim\limits_{x \to \infty} \frac{x (\ln(e+1/x)-\ln e}{\frac{1}{x}}$
and change to $t=\frac{1}{x}$, $\lim\limits_{x \to 0} \frac{\ln(\frac{e+t}{e})}{t^2}$= $\infty$
however im not sure with this result, because i got $\infty $ in third one. is my attempt correct? thanks so much!!!
For point 1, you can't do this step in this way
$$\color{red}{\alpha=\lim\limits_{x \to \infty} {\frac{f(x)}{x}}= \frac{x \log(e+\frac{1}{x}) }{x}=\log(e+\frac{1}{x})=\log e}$$
indeed you should write
$$\alpha=\lim\limits_{x \to \infty} {\frac{f(x)}{x}}= \lim\limits_{x \to \infty}\frac{x \log(e+\frac{1}{x}) }{x}=\lim\limits_{x \to \infty} \log(e+\frac{1}{x})=\log e=1$$
For point 2 note that
$$\beta=\lim\limits_{t \to 0^+}{\frac{\ln(\frac{e+t}{e})}{t}}=\lim\limits_{t \to 0^+}\frac1e\cdot{\frac{\ln(1+ \frac{t}e)}{\frac{t}e}}=\frac1e$$
For point 3 note that
$$\frac{x\log(e+\frac{1}{x}) -(x+\frac1e)}{\frac{1}{x}}=\frac{\log(e+t) -\left(1+\frac{t}{e}\right)}{t^2}=\frac{\log e+\log \left(1+\frac{t}{e}\right) -\left(1+\frac{t}{e}\right)}{t^2}=\frac{1+\log \left(1+\frac{t}{e}\right) -\left(1+\frac{t}{e}\right)}{t^2}=\frac{\log \left(1+\frac{t}{e}\right) -\frac{t}{e}}{t^2}\to-\frac1{2e^2}$$
indeed for $y\to0$
$$\log(1+y)=y-\frac{y^2}{2}+o(y^2)\implies \log \left(1+\frac{t}{e}\right) =\frac{t}{e}-\frac{t^2}{2e^2}+o(t^2)$$
and
$$\frac{\log \left(1+\frac{t}{e}\right) -\frac{t}{e}}{t^2}=\frac{\frac{t}{e}-\frac{t^2}{2e^2}+o(t^2)-\frac{t}{e}}{t^2}=-\frac1{2e^2}+o(1)\to\frac1{2e^2}$$