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Show that lim g(x)=f'(x), f'(x) = -1/x^2, g(x) = (1/(x+h)-1/x)/h
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did u try this ?where r u stuck ?
Umm I'm not really sure where to begin.
ok, so u need to find \(\large \lim \limits_{x \rightarrow 0} \dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}\) right ?
can u solve numerator part ?
yes. to show that it is equal to \[\frac{ -1 }{x ^{2} }\]
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1/x-1/x=0? so h/h, which = 1?
sorry, it should be \(\large \lim \limits_{h \rightarrow 0} \dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}\) and no, how u got that ?
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