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Let $\lambda \in \mathbb{R}$ and Let $A$ be a linear operator on $\mathbb{R}^3$. Define $f:\mathbb{R}^3 \to \mathbb{R}$ by $f(x):=\langle A(x),x\rangle- \lambda \langle x,x\rangle $ where $\langle,\rangle$ is the usual dot product on $\mathbb{R}^3$.

Does $\lim_{x\rightarrow0}\frac{f(x)}{|x|}$ exist?

My computation: $$\begin{align*} \lim_{x\rightarrow0}\frac{f(x)}{|x|} &=\lim_{x\rightarrow0}\frac{\langle A(x)-\lambda x,x\rangle}{|x|} \\ &=\lim_{x\rightarrow0}\frac{|A(x)-\lambda x||x|\cos \theta_{x}}{|x|} \\ &=\lim_{x\rightarrow0}|A(x)-\lambda x|\cos \theta_{x}=0\end{align*}$$ where $\theta_{x}$ is the angle between $A(x)-\lambda x$ and $x$. I wonder whether $\theta_{x}$ is continuous? If yes, then my computation is ok.

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up vote 5 down vote accepted

$$\left| \frac{f(x)}{||x||}\right|=\left| \frac{\langle A(x),x\rangle - \lambda \langle x,x\rangle}{||x||}\right| \leq \left| \frac{\langle A(x),x\rangle}{||x||}\right| + |\lambda|\left| \frac{\langle x,x\rangle}{||x||}\right| \leq ||Ax|| + |\lambda|\,||x||$$ $$\leq ||A||\, ||x|| + |\lambda| \,||x|| \rightarrow 0 \,,$$ as $x \rightarrow 0 \,.$

Note that, the Cauchy-Schwarz inequality has been used $$ |\langle f,g\rangle | \leq ||f||\,||g|| \,. $$

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I don't think you need $\,\cos \theta_x\,$ to be continuous: $$|Ax-\lambda x|\cos\theta_x=|(A-\lambda I)x|\cos\theta_x$$

Now, just point out that $\,|Tx|\xrightarrow [x\to 0]{}0\,$ , for any linear operator $\,T\,$ , and $\,|\cos\theta_x|\leq 1\,$ (is bounded)

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