# Using $\epsilon-\delta$ to prove that a limit is not a specific number

I want to use $\epsilon-\delta$ formulation to prove that $\lim_{x\to 1 } 2x+3 \neq 6$.

I know that I need to show that there exists an $\epsilon>0$ such that for every $\delta>0$ there exists $x\in \mathbb{R}$ such that $0<\left| x-1 \right| <\delta$ but $|2x-3|\geq \varepsilon$.

Will someone please guide me through the process?

• What happens for $x=1$ ? :-) – Hippalectryon Nov 25 '17 at 10:09
• My first approach would be to show $\lim_{x\to 1} 2x+3=5.$ Knowing that limits are unique (if they exist) in $\mathbb{R}$ with the standard topology then shows that $\lim_{x\to 1} 2x+3\not= 6.$ – Jonas Lenz Nov 25 '17 at 10:24
• How about trying something like $x=1 + \min(\frac1{10},\frac{\delta}{2})$ so $0<\left| x-1 \right| <\delta$ and $|2x-3|\geq \frac45$ – Henry Nov 25 '17 at 10:29

This problem is teaching you the negation of “for every $\epsilon > 0$ there exists a $\delta > 0$ such that ... “

So pick $\epsilon = 0.1$. Now you need to show there’s no possible $\delta$ such that, for all $x \in (1-\delta , 1 + \delta)$, $f(x)=2x+3 \in (5.9, 6.1)$. Well big values of $\delta$ will include values of $f(x)$ that are far away from 6, and small such values will never let $f(x)$ get near 6.

• Thanks a lot !!! – JosepeCorleini Nov 25 '17 at 17:37

You have a mistake in your definition.

The formal definition stats :

If for every number $ε>0$ there is some number $δ>0$ such that if :

$$|x - α| < δ$$

then

$$|f(x) - L| < ε$$

where $L= \lim_{x\to a} f(x)$.

So, for your exercise, $L=\lim_{x\to 1}f(x)=\lim_{x\to 1} (2x+3) =5$

Which translates to :

$$|x-1| < δ$$

$$|(2x+3) - 5|<ε$$

Take a look into the second expression :

$$|(2x+3)-5|<ε \Rightarrow |2x -2|<ε \Rightarrow |x-1|<\frac{ε}{2}$$

So in order to make sure that $|f(x)−5|<ϵ$, it is enough to require that $|x−1|<\frac{ε}{2}$.

Thus we can select $δ = ε/2$.

Then $δ>0$, and if $0<|x-1|<δ$, then it will follow that $|f(x)-5|<2δ=ε$

Thus, for all $\epsilon\gt 0$ there exists a $\delta\gt 0$ (namely, $\delta=\epsilon$) with the property that if $0\lt |x-1|\lt \delta$, then $|f(x)-5|\lt \epsilon$. This proves that $\lim\limits_{x\to 1}f(x) = 5\neq 6$, as desired, because limits are unique in $\mathbb R$, if they exist.

You want to find a way of making sure that the distance between $2x+3$ and $6$ is large for $x$ close enough to $1$.

Try making $2x+3\lt 5.5$, which would mean you could make $\epsilon = 0.5$.

• Thanks a lot !!! – JosepeCorleini Nov 25 '17 at 17:36