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?
Asked
Active
Viewed 886 times
1
-
1Even 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 $\lim_{x\to 0}\frac{\sin x}{x}=1$. – 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 $x\in [(x_0-\delta, x_0+\delta)\setminus{x_0}]\cap D$). – user Nov 10 '18 at 23:03
2 Answers
5
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
$$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{2x+2}{2x-2}=\dfrac{2(2)+2}{2(2)-2}=\dfrac{6}{2}=3$$
Key Flex
- 9,475
- 7
- 17
- 34
-
Thank you! Sorry, I've just seen your post , because I never srolled down. – Doesbaddel Nov 10 '18 at 22:43
-
2
-
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
-
1
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
user
- 154,566
-
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
-
1@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. – user Nov 10 '18 at 22:25
-
-
1@Doesbaddel Well done, I hope now your doubt is clear. If you are interested refer also to $\lim_{x \to 0};\frac{\sin\left(\frac 1x\right)}{\sin \left(\frac 1 x\right)}$. – user 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
-
@Doesbaddel You can change the answer to accept anytime! Just select the which one you prefer and helped you much more. If they are equivalent, in my opinion you should select the first one given, that is the one by Key Flex. Bye – user Nov 10 '18 at 22:51
-
Yeah, its hard to decide. I would suggest, that one should be allowed to accept more than one answer. I will just leave it like that and consider your tip next time I encounter two equivalently good answers. ;) – Doesbaddel Nov 10 '18 at 22:54
-