# Calculate the limit $\lim \limits_{x \to 2} \left(\frac{x^2+2x-8}{x^2-2x}\right)$, although $x\neq 2$

I should calculate the Limit $$\lim \limits_ {x \to 2} \left(\frac{x^2+2x-8}{x^2-2x}\right)$$, although I noticed, that $$x\neq 2$$ must apply. Is the limit undefined? Otherwise, with which steps should I go on to calculate the limit?

• Even there is no "cancellation", the limit $\lim_{x\to x_0}f(x)$ has nothing to do with the value of the function $f$ at $x=x_0$. For instance $f(x)=\frac{\sin x}{x}$ is not defined at $x=0$ and there is no "cancellation", but we have $\lim_{x\to 0}\frac{\sin x}{x}=1$. – user587192 Nov 10 '18 at 22:57
• @user587192 In some sense the cancellation there is also there if we consider that $\frac{\sin x}x=\frac{x+o(x)}{x}=1+o(1)\to 1$. The key point, as younoticed, is that when we take the limit we are considering $x\neq x_0$ (and more in general $x\in [(x_0-\delta, x_0+\delta)\setminus\{x_0\}]\cap D$). – gimusi Nov 10 '18 at 23:03

We have that

$$\frac{x^2+2x-8}{x^2-2x}=\frac{\color{red}{(x-2)}(x+4)}{x\color{red}{(x-2)}}=\frac{x+4}{x}$$

and then take the limit.

To clarify why we are allowed to cancel out the $$(x-2)$$ factor refer to the related

• Thank you! I didn't knew I could still simplify like that, because this will obviously fix one "gap". – Doesbaddel Nov 10 '18 at 22:24
• @Doesbaddel We can because, by defiition, when we take the limit $x\to 2$ we are considering $x\neq 2$. Refer to the given link for more details on that. – gimusi Nov 10 '18 at 22:25
• I looked it up, thank you and have a nice day – Doesbaddel Nov 10 '18 at 22:39
• @Doesbaddel Well done, I hope now your doubt is clear. If you are interested refer also to $\lim_{x \to 0}\;\frac{\sin\left(\frac 1x\right)}{\sin \left(\frac 1 x\right)}$. – gimusi Nov 10 '18 at 22:42
• Thank you man. I voted so fast, that I completely overlooked the other persons answer. Haha – Doesbaddel Nov 10 '18 at 22:47

Given $$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{\color{red}{(x-2)}(x+4)}{x\color{red}{(x-2)}}=\lim_{x\rightarrow2}\dfrac{x+4}{x}=\lim_{x\rightarrow2}1+\dfrac4x = 3$$

OR

You could also use L'Hopital's rule

$$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{2x+2}{2x-2}=\dfrac{2(2)+2}{2(2)-2}=\dfrac{6}{2}=3$$

• Thank you! Sorry, I've just seen your post , because I never srolled down. – Doesbaddel Nov 10 '18 at 22:43
• @Doesbaddel Why Sorry? Hope my posts helped you. – Key Flex Nov 10 '18 at 22:44
• I would've labeled your post as accepted otherwise. Unfortunately I was a bit too fast with the first one. – Doesbaddel Nov 10 '18 at 22:46
• Yeah, but I can't decide :(( It would be easier to accept more than one answer. – Doesbaddel Nov 10 '18 at 22:48
• @Doesbaddel Accept the answer which you feel helped you in solving the above question. – Key Flex Nov 10 '18 at 22:49