0
$\begingroup$

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?

Thanks a lot in advance

$\endgroup$
  • $\begingroup$ What happens for $x=1$ ? :-) $\endgroup$ – Hippalectryon Nov 25 '17 at 10:09
  • $\begingroup$ 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. $ $\endgroup$ – Jonas Lenz Nov 25 '17 at 10:24
  • $\begingroup$ 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$ $\endgroup$ – Henry Nov 25 '17 at 10:29
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thanks a lot !!! $\endgroup$ – JosepeCorleini Nov 25 '17 at 17:37
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\epsilon = 0.5$.

$\endgroup$
  • $\begingroup$ Thanks a lot !!! $\endgroup$ – JosepeCorleini Nov 25 '17 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.