Calculate $\lim\limits_{x \to 0}{\frac{||a + xb|| - ||a||}{x}}$.

Question

Non-zero vectors a, b, s.t. $||b||=1$, angle of $a$ and $b$ is $\pi/4$.

Calculate the limit: $\lim\limits_{x \to 0}{\frac{||a + xb|| - ||a||}{x}}$.

Answer 1

Note that \begin{align} \dfrac{\Vert a+xb \Vert - \Vert a \Vert}x & = \dfrac{\Vert a+xb \Vert^2 - \Vert a \Vert^2}{x \left(\Vert a+xb \Vert + \Vert a \Vert\right)}\\ & = \dfrac{\Vert a \Vert^2 +2x \vec{a} \cdot \vec{b} + x^2 \Vert b \Vert^2 - \Vert a \Vert^2}{x \left(\Vert a+xb \Vert + \Vert a \Vert\right)}\\ & = \dfrac{2x \vec{a} \cdot \vec{b} + x^2 \Vert b \Vert^2}{x \left(\Vert a+xb \Vert + \Vert a \Vert\right)}\\ & = \dfrac{2 \vec{a} \cdot \vec{b} + x \Vert b \Vert^2}{\left(\Vert a+xb \Vert + \Vert a \Vert\right)}\\ \end{align} Now let $x \to 0$ and conclude the limit.