Is the limit for this function exist?

$\lim_{ x\to 3}$ for $f(x)=|2x-4|$

I think the limit for this function is $2$.

Is my answer correct or not? Because my teacher said there is no limit for this function. So I am confused


closed as off-topic by vonbrand, Thomas, Shailesh, Leucippus, choco_addicted Apr 5 '16 at 1:26

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  • 4
    $\begingroup$ I dont see any reason why your answer should be wrong. The function is continous and the limit is well defined. $\endgroup$ – MrYouMath Apr 4 '16 at 16:30
  • $\begingroup$ @MrYouMath that means my answer is correct, right? $\endgroup$ – Tony Apr 4 '16 at 16:33
  • 1
    $\begingroup$ Yes, your answer is correct. $\endgroup$ – MrYouMath Apr 4 '16 at 16:34

\begin{align} \lim_{x\to 3} |2x-4| &= \left|\lim_{x \to 3}(2x-4)\right|\\[0.3cm] &= \left|2(3)-4\right|\\[0.3cm] &= 2 \end{align}

We can push the limit into the absolute value because the absolute value function is continuous and it's ok to push limits into continuous functions.

So $|2x-4|$ does have a limit which exists as $x \to 3$.

A related statement that's similar to what you were told (except for the fact that this related statement is true) is that $|2x-4|$ doesn't have a derivative at $x=2$. Note that that's 2 and not 3.


Notice that $|2x-4|=2x-4$ when $x\ge2$ and then $$\lim_{x\to3}|2x-4|=\lim_{x\to3}(2x-4)=2(3)-4=2$$


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