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lim(x->0) [(1-2x^2)^1/5-(1+3x^2)^1/3]/x^2
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or, \[\Large \lim_{x \rightarrow 0}~ \left(\begin{matrix} \\ \frac{(1-2x^2)^{1/5}-(1+3x^2)^{1/3}}{x^2}\end{matrix}\right)\]
second
what can we solve without L'H'S?
there certainly are some things that we don't have to apply L'H"S to...
no, actually, nvm, just do L'H'S. I am sorry for confusing you. \[\Large \lim_{x \rightarrow 0}~ \left(\begin{matrix} \\ \frac{(1-2x^2)^{1/5}-(1+3x^2)^{1/3}}{x^2}\end{matrix}\right)\]
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it will involve a chain rule for each of the tops, for (1-2x^2)^1/5 and for the second one. then we will need to simplify it. can you first tell me what the derivative of (1-2x^2)^1/5) is ?
1/5*(1-2x^2)^-4/5*-4x
|dw:1417536111828:dw|
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