# Find limit from first principle

Find the limit or prove the limit does not exist using the definition of the limit:

$$\lim\limits_{x \rightarrow c} x^2+x+1.$$

I am getting stuck in the problem following through on the algebra to figure out a $\delta$ to choose.

• Here is the technique. – Mhenni Benghorbal Jun 27 '13 at 15:38
• @MhenniBenghorbal You would find that $|f(x)-L|<M|x-c|$ (e.g.) and you choose $\displaystyle \delta=\min\left\{\frac{\varepsilon}{M},1\right\}$. While I understand that this is the fully correct way to proceed (for example if $\varepsilon$ is large), I usually just say that if we are looking at $x\rightarrow c$ then we can always have $\delta<1$. What do you think of that approach... i.e. I just take $\delta=\varepsilon/M$? – JP McCarthy Jun 27 '13 at 15:48
• Then you don't end up proving what you want, unless you specify from the beginning something like "wlog, we can assume $\varepsilon \leq M$". – Clement C. Jun 27 '13 at 15:50

Fix any $\varepsilon > 0$. Define $M=|c|+1$, and $\delta = \min\left(\frac{\varepsilon}{2M},1\right)$. Then, for any $x$ such that $|x-c|\leq \delta$, \begin{align*} |c^2+c+1-(x^2+x+1)|&=|c-x||c+x+1| \leq \delta (|c|+1+|x|) \\ &\leq \delta\cdot (M+|c|+\delta) \\ &\leq \delta\cdot (M+|c|+1) = \delta\cdot 2M \\ &\leq \varepsilon \end{align*}
• If you do the proof without replacing anything by $M$ (just assuming $\delta \leq 1$), you end up with $\delta\cdot 2(|c|+1)$. In hindsight, that motivates setting a parameter $M=|c|+1$ to simplify the expressions. – Clement C. Jun 27 '13 at 15:56
• You are trying to show that for all $\epsilon>0$ that you can find a $\delta>0$ such that if $|x-c| < \delta$, then $|f(c)-f(x)| < \epsilon$. Here $f(c)=c^2+c+1$. – copper.hat Jun 27 '13 at 16:02
• @user72195: the only possible limit "that'd make sense" is $c^2+c+1$, so I set up to prove that $x^2+x+1\to c^2+c+1$. As a matter of fact, more generally, for any continuous function $f$, you'll have $f(x)\to f(c)$. – Clement C. Jun 27 '13 at 16:03