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The following expression:

$$\frac{\sqrt{4+h}-2}{h}$$

should be simplified to:

$$\frac{1}{\sqrt{4+h}+2}$$

While looking at this site I found this already answered, but my problem is why does the answer contain a radical in the denominator when that is against the rules. or is it? isn't the first form the simplest form? I couldn't ask this question on the original post.

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The second form is considered the simplified form in this case because it does not contain the lone $h$ of the original form. This type of expression comes from a difference quotient. For a function $f(x)$ at a point $a$, we define the difference quotient as $\dfrac{f(a+h) - f(a)}{h}$ (note that there are alternate but equivalent definitions). For the above problem, it appears $f(x) = \sqrt{x}$ and $a=4$. When working with these expressions in precalculus, the main goal is to simplify them so that the $h$ that started in the denominator is eliminated. This is preparation for calculus where you will want to substitute $0$ for $h$, which can not be done in the original form.

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    $\begingroup$ It throws me off because when I see it in the latter form, I think I need to get rid of the radicals, I didn't even think of the difference quotient at all, though looking at the original, it seems to be already in the right format (assuming h doesn't equal 0). Any advice on what to review to recognize this type of problem? $\endgroup$ – Joshhw Jan 31 '14 at 16:35
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    $\begingroup$ At the precalc level, the problem statements should generally be simplify. So if you are told to simplify a fraction with a lone $h$ in the denominator and functions having $h$ in numerator, it is a tip you need to simplify to eliminate the $h$ on the bottom. $\endgroup$ – John Habert Jan 31 '14 at 16:39
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The given expression on its first form is undefined at $0$ and its limit at $0$ has an indeterminate form $\frac 0 0$ and to avoid this problem we write the expression on its second form which's defined at $0$ and the limit can be evaluated.

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In calculus, to find the derivative using the definition of the derivative, you would need to evaluate that expression at $h = 0$. However, in the original form, this is impossible because it gives you $\frac{0}{0}$. This usually means that your expression, when evaluate at $h = 0$, is in the form of $\frac{0 \times\text{ stuff}}{0 \times\text{ other stuff}}$, and by "simplifying" the expression, you can get rid of the 0 term multiplying both sides and properly evaluate the expression.

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