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I'm stumped on yet another assignment problem. I'm not allow to use power rule with this problem so i have to rely on good old $$ \frac{f(a+h)-f(a)}{h} $$

so here are the steps ive taken thus far but i cant quite bring it home. 1- $$\lim_{h\to 0}\frac{\frac{1}{\sqrt{t+h}}- \frac{1}{\sqrt{t}}}{h} $$

2- get common denominator $ \sqrt{t} \sqrt{t+h} $ $$\lim_{h\to 0}\frac{\frac{\sqrt{t}}{\sqrt{t+h}} - \frac{\sqrt{t+h}}{\sqrt{t}}}{h} $$

3- multiply by conjugate pair $$\lim_{h\to 0}\frac{\frac{\sqrt{t}}{\sqrt{t+h}} - \frac{\sqrt{t+h}}{\sqrt{t}}}{h}* \frac{\sqrt{t}+\sqrt{t+h}}{\sqrt{t}+\sqrt{t+h}} $$

4-multiply across and cancel the h's and i end up with $$ \frac{-1}{\sqrt{t+h}\sqrt{t}(\sqrt{t}+\sqrt{t+h} )}$$

this is where im stuck the solutions manual gets to $\frac{-1}{\sqrt{t}\sqrt{t}(\sqrt{t}+\sqrt{t})} $

i have no idea how they could have achieved it? I'm missing an intermediate step can someone please point me in the right direction and i think my algebra is failing me here.

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    $\begingroup$ You're not stuck, you're done! Let $h$ go to $0$, and ... $\endgroup$ – user452 Feb 24 '13 at 16:47
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    $\begingroup$ $\sqrt{\cdot}$ is continuous. $\endgroup$ – Stefan Feb 24 '13 at 16:47
  • $\begingroup$ @ trb456 thank you now i get i!!! i feel like a moron now! thank you for pointing that out to me. $\endgroup$ – Miguel Feb 24 '13 at 16:58
  • $\begingroup$ Your step (2) is wrong, you don't add the fractions. $\endgroup$ – vonbrand Feb 24 '13 at 17:20
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Since the square root function is continuous, then you can let the limit "pass through" the radicals, and take $h$ to $0$. We still have to be a bit careful, though. What value(s) of $t$ in the original domain will cause issues in the resulting expression?

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  • $\begingroup$ would that be 0 right? $\endgroup$ – Miguel Feb 24 '13 at 16:59
  • $\begingroup$ I'm not sure that is allowed by OP's current mathematical arsenal. $\endgroup$ – vonbrand Feb 24 '13 at 17:21
  • $\begingroup$ @Miguel: Yes, indeed. So, while the original domain is all nonnegative numbers, the domain of the derivative is only the positive numbers. Have you seen the result that we can let limits "pass through" continuous functions before? $\endgroup$ – Cameron Buie Feb 24 '13 at 18:02

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