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lim(h→0) ((1+h)^2+1-(1^2+1))/h)
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this is the derivative of \[x^2\] at x = 1 so you should get 2
quite not .. though answer would be 2
\[(1+h)^2+1=1^2+2h+h^2+1=2+2h+h^2\] \[1^2+1=2\] \[2+2h+h^2-2=2h+h^2\] \[\frac{2h+h^2}{h}=\frac{h(2+h)}{h}=2+h\]
i guess x^2+1
yes you are right it would be the derivative of \[x^2+1\] at x = 1
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thanx
yw
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