Simple looking matrix limit I am trying to evaluate:
$$\lim_{V\rightarrow {\bf{0}}^{n\times n}}\frac{|\text{tr}(V^2)|}{\|V\|_{op}}$$
($\text{tr}(V^2)$ is the trace of the matrix $V^2$)
Hoping that the limit is $0$ to show a function is Frechet differentiable, it seems this should be the answer intuitively.
I am slightly stuck, my only idea at the moment is 
$$\lim_{V\rightarrow {\bf{0}}^{n\times n}}{|\frac{1}{\|V\|_{op}}\text{tr}(VV)|} = \lim_{V\rightarrow {\bf{0}}^{n\times n}}{|\text{tr}(\hat{V}V)|} = \text{tr}({\bf{0}}^{n\times n}) = 0$$
Which feels incorrect.
 A: Assume that $V$ approaches the zero matrix from a constant direction $X$, i.e.  $$V=\beta X$$
Then
$$\eqalign{
 L &= \lim_{V\to 0} \frac{|{\rm tr}(V^2)|}{\|V\|} \cr 
 &= \lim_{\beta\to 0} \frac{|\beta|^2\,\,|{\rm tr}(X^2)|}{|\beta|\,\,\|X\|} \cr
 &= \frac{|{\rm tr}(X^2)|}{\|X\|}\,\,\lim_{\beta\to 0}|\beta| \cr
 &= 0 \cr
}$$
A: Write $V = [v_{ij}]$ and notice that by the Cacuchy-Schwarz inequality
$$ \left| \operatorname{tr}(V^2) \right|
= \left|\sum_{i,j=1}^{n} v_{ij}v_{ji} \right|
\leq \sum_{i,j=1}^{n} |v_{ij}|^2
= \operatorname{tr}(V^* V)
= \sum_{i=1}^{n} \lambda_i $$
where $V^*$ is the adjoint (conjugate transpose) of $V$ and $\lambda_i$'s are eigenvalues of $V^*V$. (Since $V^*V$ is positive-semidefinite Hermitian matrix, it follows that $\lambda_i \geq 0$.) We also notice that
$$ \|Vx\|^2 = \langle Vx, Vx\rangle = \langle x, V^* Vx \rangle $$
and hence by the spectral theorem we obtain
$$ \| V \|_{\text{op}}
= \max_{\|x\|=1} \frac{\|Vx\|}{\|x\|}
= \left( \max_{i=1,\cdots,n} \lambda_i \right)^{1/2}. $$
Combining altogether,
$$ \left| \operatorname{tr}(V^2) \right|
= \sum_{i=1}^{n} \lambda_i
\leq n \left( \max_{i=1,\cdots,n} \lambda_i \right)
= n \| V\|_{\text{op}}^2. $$
Therefore the conclusion follows from the squeezing lemma.

Let me also give a slightly different reasoning: Consider the Frobenius norm
$$\| V \|_F = \sqrt{\operatorname{tr}(V^*V)} = \left( \sum_{i,j=1}^{n} |v_{ij}|^2 \right)^{1/2}.$$
It is easy to check that $ (A, B) \mapsto \operatorname{tr}(A^* B) $ is an inner product and hence $\|V\|_F$ is indeed a norm. Now recall that any two norms on a finite-dimensional space are comparable to each other, so there exists an absolute constant $C > 0$ for which $ \| V\|_F \leq C \|V\|_{\text{op}} $ holds for any $n\times n$ matrix $V$. So by the Cauchy-Schwarz inequality,
$$ \left|\operatorname{tr}(V^2)\right| \leq \| V^* \|_F \| V \|_F = \| V \|_F^2 \leq C^2 \|V \|_{\text{op}}^2 $$
and again the limit is zero by the squeezing lemma.
