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I'm trying to solve a problem with two variables, but the limit only addresses one. Wolfram Alpha simplifies it for h. Should I leave it simplified, or does x need to be a real number?

$$\lim_{h\to 0} \frac{\sqrt{3x+h}-\sqrt{3x}}{h}$$

Wolfram Alpha's solution.

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    $\begingroup$ You don't need to use Wolfram Alpha here. Multiply by $$\frac{\sqrt{3x+h}+\sqrt{3x}}{\sqrt{3x+h}+\sqrt{3x}}$$ $\endgroup$ – oldrinb Sep 8 '13 at 21:08
  • $\begingroup$ What limit exactly are you trying to compute?You didn't mention any variable for the limit. $\endgroup$ – Git Gud Sep 8 '13 at 21:09
  • $\begingroup$ @oldrinb that's not what I mean. I meant whether or not I could simplify h. $\endgroup$ – Mia Sep 8 '13 at 21:09
  • $\begingroup$ @GitGud he did, ugly notation though... $\endgroup$ – AlexR Sep 8 '13 at 21:09
  • $\begingroup$ @MiaCcio what does it mean to 'simplify $h$'? $\endgroup$ – oldrinb Sep 8 '13 at 21:10
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If the limit is as $h$ approaches 0, then the $x$ is not changing, only $h$ is. So, $x$ is treated as if it were a constant in the calculation.

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Can you put the limit in the form $ \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $?

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Hint: $$ \begin{align} \lim_{h\to0}\frac{\sqrt{3x+h}-\sqrt{3x}}{h} &=\lim_{h\to0}\frac{\sqrt{3x+h}-\sqrt{3x}}{h}\frac{\sqrt{3x+h}+\sqrt{3x}}{\sqrt{3x+h}+\sqrt{3x}}\\ &=\lim_{h\to0}\frac{(3x+h)-(3x)}{h(\sqrt{3x+h}+\sqrt{3x})} \end{align} $$

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