Answer
See explanations.
Work Step by Step
Step 1. We can identify the quantities as: $f(x)=\frac{1}{x^2},c=\sqrt 3,L=\frac{1}{3}$.
Step 2. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ when $0\lt |x-\sqrt 3|\lt\delta$, we have $|\frac{1}{x^2}-\frac{1}{3}|\lt\epsilon$
Step 3. The last inequality gives $-\epsilon\lt \frac{1}{x^2}-\frac{1}{3}\lt\epsilon$ or $\frac{1}{3}-\epsilon\lt \frac{1}{x^2}\lt\frac{1}{3}+\epsilon$
Step 4. Assume $\epsilon\lt\frac{1}{3}$, and $x\gt0$; we can get $\frac{1}{\frac{1}{3}+\epsilon}\lt x^2 \lt \frac{1}{\frac{1}{3}-\epsilon}$ and $\sqrt {\frac{1}{\frac{1}{3}+\epsilon}} \lt x \lt \sqrt {\frac{1}{\frac{1}{3}-\epsilon}}$
Step 5. From the requirement for the value $\delta$ in step 2, we have $-\delta\lt x-\sqrt 3\lt\delta$ or $\sqrt 3-\delta\lt x\lt\sqrt 3+\delta$
Step 6. Compare the results from steps 4 and 5, let $\sqrt 3-\delta=\sqrt {\frac{1}{\frac{1}{3}+\epsilon}}$, we get $\delta=\sqrt 3-\sqrt {\frac{1}{\frac{1}{3}+\epsilon}}$. Let $\sqrt 3+\delta=\sqrt {\frac{1}{\frac{1}{3}-\epsilon}}$, we get $\delta=\sqrt {\frac{1}{\frac{1}{3}-\epsilon}}-\sqrt 3$. Thus, we can set $\delta$ as the smaller value of $\sqrt 3-\sqrt {\frac{1}{\frac{1}{3}+\epsilon}}$ and $\sqrt {\frac{1}{\frac{1}{3}-\epsilon}}-\sqrt 3$
Step 7. For $\epsilon\geq\frac{1}{3}$, choose $\delta$ to be the smaller of $1$ and $\sqrt 3-\sqrt {\frac{1}{\frac{1}{3}+\epsilon}}$
Step 8. With the setting of the $\delta$ value, we can go backwards through the above steps to reach a conclusion thatfor all $x$ when $0\lt |x-\sqrt 3|\lt\delta$, we have $|\frac{1}{x^2}-\frac{1}{3}|\lt\epsilon$ which proves the limit statement.