cj104
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I managed to find the answer using uv'-vu'/v^2 but I wanna know how we can find the answer using limits with something like f(x+h)-f(x)/h
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RDKGames
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(Original post by cj104)
I managed to find the answer using uv'-vu'/v^2 but I wanna know how we can find the answer using limits with something like f(x+h)-f(x)/h
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\displaystyle m = \lim_{h \to 0} \dfrac{\frac{3}{\sqrt{a+h}} - \frac{3}{\sqrt{a}}}{h} = \lim_{h \to 0}\dfrac{3\sqrt{a} - 3\sqrt{a+h}}{h\sqrt{a}\sqrt{a+h}  }

It is now useful to use Binomial theorem on \sqrt{a+h} in increasing powers of h since h is small when it tends to zero.


EDIT: Even better, just multiply numerator and denominator by 3\sqrt{a} + 3\sqrt{a+h}, simplify to get rid of the singularity at h=0, and get the limit by subbing in h=0.
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DFranklin
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(Original post by RDKGames)
EDIT: Even better, just multiply numerator and denominator by 3\sqrt{a} + 3\sqrt{a+h}, simplify to get rid of the singularity at h=0, and get the limit by subbing in h=0.
Not only is this simpler, but it avoids the circularity that proofs of the binomial theorem usually assume you've already proved (1+x)^\alpha is differentiable.
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