Prove that if $f$ differentiable in $x^*$ and $f(x^*)=0$, then $\liminf_{x\to x^*}\frac{|f(x)|}{||x-x^*||}=0$ if $n>1$ Let $f\colon\mathbb{R}^n\to\mathbb{R}$ be differentiable at $x^*$ and $f(x^*)=0$. Prove that if $n>1$, then $$\liminf_{x\to x^*}\frac{|f(x)|}{||x-x^*||}=0.$$
Is this true for $n=1$?
I know that because $f$ is differentiable in $x^*$, there is a linear transformation $L\colon \mathbb{R}^n\to\mathbb{R}$ such that $$0=\lim_{x\to x^*}\frac{|f(x)-f(x^*)-L(x-x^*)|}{||x-x^*||}=\lim_{x\to x^*}\frac{|f(x)-L(x-x^*)|}{||x-x^*||}.$$ How to proceed from there? And what is special about $n=1$?

Why the difference between the cases $n=1$ and $n>1$?
For $n=1$, the kernel of a linear map can be reduced to the zero vector. This isn't the case for $n>1$.
The result is true for $n>1$
In that case $L$ is a linear form and its kernel is not reduced to the zero vector. Take $a \in \ker L \setminus \{0\}$. For $m \in \mathbb N$ and $x_m = \frac{a}{m} + x^*$, you have $L(x_m-x^*) = 0$ and $\lim\limits_{m \to \infty} \dfrac{|f(x_m)-L(x_m-x^*)|}{||x_m-x^*||}= \dfrac{|f(x_m)|}{||x_m-x^*||}=0$. As $\lim\limits_{m \to \infty} x_m = x^*$ you have $\lim \inf_{x\to x^*}\dfrac{|f(x)|}{||x-x*||}=0$.
The result is wrong for $n=1$
Just take $f(x)=x$ and $x^*=0$. You have $\dfrac{|f(x)|}{||x-x*||}=1$ for all $x \in \mathbb R \setminus \{0\}$.
