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Can someone explain why these two limits are equal?

$$\lim_{h\to 0} \frac{(5+h)^2-25}{h} = \lim_{h\to 0} \frac{(5+h+5)(5+h-5)}{h}$$

Why I multiply out the second part I get $10h + h^2$ and I don't see how that's equal to $(5+h)^2$.

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    $\begingroup$ It is equal to $(5+h)^2-25$, which was claimed. $\endgroup$ – Dietrich Burde Feb 22 '16 at 16:23
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    $\begingroup$ $(5+h-5)(5+h+5)=25+5h-25+5h+h^2-5h+25+5h-25=5h+5h+h^2=2(5h)+h^2$ and $(5+h)^2-25=25 +2(5h)+h^2-25= 2(5h)+h^2$. $\endgroup$ – Mauro ALLEGRANZA Feb 22 '16 at 16:29
  • $\begingroup$ Why should it be? The first part isn't $(5+h)^2$. The first part is $(5+h)^2 -25$. -25 is not the same thing as 0. $10h + h^2$ is equal to $(5+h)^2 - 25$. $\endgroup$ – fleablood Feb 22 '16 at 16:55
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Given $a,b\in\mathbb{R}$, you have the well-known formula for the difference of two squares, which is given by $$a^2-b^2 = (a+b)(a-b).$$ In your case, take $a=5+h$ and $b=5$.

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Well, I don't know how people can forget the well known formula $a^2-b^2=(a+b)(a-b)$ However $(5+h)^2-25=25+10h+h^2-25=10h+h^2$ and $(5+h+5)(5+h-5)=10h+h^2$
So they are equal

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For the first limit:

$$\lim\limits_{h\to 0} \frac{(h+5)^2-25}{h}=\lim\limits_{h\to 0} \frac{h^2+10h+25-25}{h}=\lim\limits_{h\to 0}\frac{h(h+10)}{h}=\lim\limits_{h\to 0} (h+10)=0+10=10$$

And for the second:

$$\lim_{h\to 0} \frac{(5+h+5)(5+h-5)}{h}= \lim_{h\to 0} \frac{(10+h)h}{h}= \lim_{h\to 0} (h+10)=0+10=10$$

And note that $$x^2-y^2 = (x+y)(x-y)$$

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