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Question Number 162539 by LEKOUMA last updated on 30/Dec/21

$$Calculate$$$$\underset{h\to 0}{\lim }\frac{f(3-h)-f\left(3\right)}{2h},with{f}^{\prime }\left(3\right)=2$$

Answered by Ar Brandon last updated on 30/Dec/21

$$\underset{h\to 0}{\lim }\frac{f(3-h)-f\left(3\right)}{2h}=\frac{0}{0}\to LH$$$$=\underset{h\to 0}{\lim }\frac{-f{}^{\prime }(3-h)}{2}=\frac{-f{}^{\prime }\left(3\right)}{2}=-1$$

Answered by mr W last updated on 30/Dec/21

$$\underset{h\to 0}{\lim }\frac{f(3-h)-f\left(3\right)}{2h}$$$$=-\frac{1}{2}\underset{h\to 0}{\lim }\frac{f(3-h)-f\left(3\right)}{(-h)}$$$$=-\frac{1}{2}\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(3+\mathrm{\Delta }x)-f\left(3\right)}{\mathrm{\Delta }x}$$$$=-\frac{1}{2}{f}^{\prime }\left(3\right)$$$$=-\frac{1}{2}\times 2$$$$=-1$$

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