User:Adaneth Mirimë/LaTeX

$$ \text{Let }G=V\ \cup\ E\text{ be a planar graph. Then, }G\text{ has chromatic number }\chi(G)\leq4. $$

Let $$G=V\cup E$$ be a planar graph. Then, $$G$$ has chromatic number $$\chi(G)\leq4$$.

Fundamental Theorem of Line Integrals: $$\int_C{\vec{F}\cdot d\vec{r}}=\int_a^b{\vec{F}(\vec{r}(t))\cdot\vec{r}'(t)dt}=\int_a^b{\nabla G(\vec{r}(t))\cdot\vec{r}'(t)dt}=G(\vec{r}(b))-G(\vec{r}(a))$$

Binomial Expansion Formula: $$(x+y)^n=\textstyle\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k$$

Prove: $$\lim_{x\to3}\left(\tfrac{4}{2x-5}\right)=4$$.

Proof: We must show $$\forall\ \epsilon>0,\ \exists\ \delta>0$$ such that $$\left|\left(\tfrac{4}{2x-5}\right)-4\right|<\epsilon$$ when $$|x-3|<\delta$$. Let $$\epsilon>0$$ and consider $$\delta\leq\min\left\{\tfrac{1}{4},\ \tfrac{\epsilon}{16}\right\}$$ with $$|x-3|<\delta$$. Then,

\begin{align} \left|\left(\tfrac{4}{2x-5}\right)-4\right|&=\left|\tfrac{4-4(2x-5)}{2x-5}\right|\\ &=\left|\tfrac{-8x+24}{2x-5}\right|\\ &=\left|\tfrac{-8(x-3)}{2x-5}\right|\\ &=\tfrac{8|x-3|}{|2x-5|}\\ &<8\cdot\delta\cdot\left(\tfrac{1}{|2x-5|}\right)\text{ since }|x-3|<\delta\\ &\leq8\cdot\delta\cdot2\text{ when }\delta\leq\tfrac{1}{4}\\ &=16\cdot\delta\\ &\leq16\cdot\left(\tfrac{\epsilon}{16}\right)\text{ when }\delta\leq\tfrac{\epsilon}{16}\\ &=\epsilon \end{align} $$

$$\therefore\lim_{x\to3}\left(\tfrac{4}{2x-5}\right)=4\ \Box$$