#Cant do either part, can someone please help!!!!

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shell glacier
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since h->0 you can replace the 2(x+h) + 1 with (2x+1) in the denominator and then cancel out the h’s to get the answer

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for b just ignore the constants and differentiate (2x+1)^(-k-2) and then you’ll get extra multiples of (k+1), 2, and -1 which you can use to get the k!, 2^k, (-1)^k into (k+1)!, 2^(k+1) and (-1)^(k+1)