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limit as x approaches 0 of... (sqrt(4+x)-2)/x = ?
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Use L'Hopital's rule. Take the derivative of the top and the derivative of the bottom.
\[\lim_{x \rightarrow 0} 1/(2\sqrt{4+x})\]
Then substitute 0 for x. Sound good?
Have you heard of L'Hopital's rule?
Nope, I hadn't. I'm trying to take the derivative at the moment, did you use the u*v' - v*u' / v^2 form, or taking the derivatives individually?
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For L'Hopital's (which is really cool), you just take the derivative of the top and put it on the top, then take the derivative on the bottom and put it on the bottom. The limit is the same, no quotient rule necessary!
So yes, I'm just taking the derivatives individually.
Oh, gotcha. Thanks :D I got it.
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