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I'm trying to understand the process that is taken to achieve the answer for the following: $$\lim_{h\to 0}\frac{\dfrac 2{a+h}-\dfrac 2a}{h}$$

I know that the answer is $-\dfrac{2}{a^2}$ , but whenever I make the denominators common and simplify everything, I end up with $\dfrac{2h^2+2ah}{2a^2+2ah}$ or $\dfrac{2h^2(a+h)}{2a^2(a+h)}$ , and I don't know where I'm going wrong at. If I simplify further, and cancel out further, then I'm left with $\dfrac {2h}{2a}$, which equals 0 as $h$ approaches 0. But that answer isn't correct.

najayaz
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cbenn95
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  • please fix the mathjax – jimjim Dec 18 '15 at 03:29
  • i'm not fluent in mathjax, so I don't really know how to fix it. there's supposed to be (a/b +c/d)/e and I can't get it to look like that. plus I've already spent like 30 minutes trying to get it to work and haven't been able to get it to look correctly, so I'm going back to studying until this is the only thing left. – cbenn95 Dec 18 '15 at 03:41
  • @just give me the a, b, c ,d,e separately and I put it together, I had the same problem as you, couldnt separate them – jimjim Dec 18 '15 at 03:51
  • $$\frac{2}{a+h} - \frac{2}{a} = \frac{2a}{a(a+h)} - \frac{2(a+h)}{a(a+h)} = \frac{2a - 2(a+h)}{a(a+h)}$$ – Daniel Fischer Dec 18 '15 at 10:47

2 Answers2

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Here's how I would calculate the limit. First off make the denominators same: \begin{align} L&=\lim_{h\to 0}\frac{2a-2(a+h)}{ah(a+h)}\\\ &=\lim_{h\to 0}\frac{-2h}{ah(a+h)}\\\ &=\lim_{h \to 0}\frac{-2}{a^2+ah}\\\ &=-\frac{2}{a^2} \end{align}

najayaz
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    Yes! Finally learned how to align equations! – najayaz Dec 18 '15 at 10:52
  • Okay I think I understand what happened here. I overlooked the (a+h) part and forgot that if you're adding you can solve for limit as long as the bottom doesn't equal 0. Thanks! – cbenn95 Dec 18 '15 at 13:01
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    Also, Thank you for fixing the mathjax coding for me, i'm gonna take a look at it in a bit to get a feel for how it should look like for complicated functions like this for future reference – cbenn95 Dec 18 '15 at 13:03
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$$\lim_{h\to 0}\frac{\dfrac 2{a+h}-\dfrac 2a}{h}=\lim_{h\to 0} \frac{-2h}{h(a^2+ah)}=\frac{-2}{a^2}$$

Kushal Bhuyan
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