Taylor Series for $x^2 \ln(1 + \frac{1}{x})$ at $x = 0$.

Question

We know that the Taylor Series of a function, grade $n$ at $x = a$ is given by $$ P_{n,a}(x) = a_0 + a_1(x-a) + a_2(x-a)^2 + ... + a_n(x-a)^n $$ where $ a_i = \frac{f^{(i)}(a)}{i!} $.

The problem that I have is that when I try to make the first term of the Taylor Series of $f(x) = x^2 \ln(1+\frac{1}{x})$, I get $$ \frac{0^2 \ln(1+\frac{1}{0})}{0!} $$ but that I can't do $\frac{1}{0}$.

Then, I have to use its Taylor Series to evaluate $$ \lim_{x \to 0} [x - x^2\ln(1+\frac{1}{x})]$$

Answer 1

hint

For $x>0$,

$$\ln(1+\frac 1x)=\ln(1+x)-\ln(x)$$ and

$$\ln(1+x)=x+x\epsilon(x)$$ with $$\lim_{x\to 0}\epsilon(x)=0$$ and $$\lim_{x\to 0^+}x\ln(x)=0$$

Answer 2

Remember that $ x \ln x \rightarrow 0$ for $x \rightarrow 0+$. Now, close to the origin

$$ f(x) = x^2(\ln(x+1) -\ln(x)) \approx x^3 -x^2 \ln x $$

This is sufficient to say that $f \rightarrow 0$. Therefore, also your last limit is zero.

Answer 3

In this answer, the power series for $\log(1+x)$ is used to show that $$ \lim_{x\to0^+}x\log(x)=0\tag1 $$ Therefore, since $\lim\limits_{x\to0}x=0$ and $\lim\limits_{x\to0}\log(1+x)=0$, $$ \begin{align} \lim_{x\to0^+}x^2\log\left(1+\frac1x\right) &=\lim_{x\to0^+}x^2\left(\log(1+x)-\log(x)\right)\tag2\\ &=\lim_{x\to0}x^2\log(1+x)-\lim_{x\to0^+}x^2\log(x)\tag3\\ &=0-0\tag4 \end{align} $$ However, the second derivative does not exist at $x=0$: $$ \frac{\mathrm{d}^2}{\mathrm{d}x^2}x^2\log\left(1+\frac1x\right) =2\log\left(1+\frac1x\right)-\frac{3+2x}{(1+x)^2} $$ so there can't be a Taylor expansion there.