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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?

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    $\begingroup$ 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 $\lim_{x\to 0}\frac{\sin x}{x}=1$. $\endgroup$ – user587192 Nov 10 '18 at 22:57
  • $\begingroup$ @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 $x\in [(x_0-\delta, x_0+\delta)\setminus\{x_0\}]\cap D$). $\endgroup$ – gimusi Nov 10 '18 at 23:03
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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

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

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

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

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