Limit in n-dimensions with dot product

Question

Let there be three vectors $\mathbf a$, $\mathbf b$, and $\mathbf h$ in $\mathbb R^n$. Define the real number $$p(\mathbf h) \doteq (\mathbf a \cdot \mathbf h) (\mathbf b \cdot \mathbf h)$$ Does the following limit exist? In particular, does it equal $0$? $$\lim_{\mathbf h \to \mathbf 0} \frac {p(\mathbf h)} {|\mathbf h|} $$ I was thinking of expressing $p$ as a product of sums, like $$p(\mathbf h)=\left(\sum_{j=1}^n a_jh_j\right)\left(\sum_{k=1}^n b_kh_k\right) = \sum_{i=1}^na_ib_ih_i^2 + \sum_{j\neq k} a_jb_kh_jh_k $$ Are there any dot product properties I may be forgetting at this point?

Answer 1

Observe that using Cauchy Schwarz Inequality yields:

$$0 \leq \bigg | \frac {p(h)} h \bigg | \leq \frac 1 {|h|} |a||b||h|^2 = |a||b||h| \to 0$$ as $h \to 0$. Hence by the Squeeze Thm your limit equals 0.

Answer 2

Observe that

$$p(h):=(a\cdot h)(b\cdot h)=\left\|a\right\|\left\|b\right\|\left\|h\right\|^2\cos\theta\cos\eta$$

with $\;\theta=\;$ the angle between $\;a,h\;$ , and $\;\eta=$ the angle between $\;b,h\;$ , and thus

$$\frac{p(h)}{\left\|h\right\|}=\left\|a\right\|\left\|b\right\|\left\|h\right\|\cos\theta\cos\eta\xrightarrow[h\to0]{}0$$

Answer 3

Using Cauchy-Schwarz inequality, we have $$ 0 \le \frac{|p(h)|}{|h|}\le \frac{|a||b||h|^2}{|h|} = |a||b||h| $$ that goes to zero as long as $h$ tends to zero.