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Evaluate $$\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x^{\frac{1}{2}}-2^{\frac{1}{2}}} \quad$$ If $$f^{\prime}(2)=4$$

$$\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x^{1 / 2}-2^{1 / 2}}$$

$$=\lim _{x \rightarrow 2} \frac{(f(x)-f(2))(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}$$

$$=\lim _{x \rightarrow 2} \frac{(f(x)-f(2))(\sqrt{x}+\sqrt{2})}{x-2}$$

$$=\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \cdot \lim _{x \rightarrow 2}(\sqrt{x}+\sqrt{2})$$

$$=F^{\prime}(2) \cdot \lim _{x \rightarrow 2} \sqrt{x}+\sqrt{2}=4 \cdot(\sqrt{2}+\sqrt{2})=4(2 \sqrt{2})=8\sqrt{2}$$

Use the definition of the derivative to find $$F^{\prime}(x),$$ where $$f(x)=\frac{1}{x-1}, x \neq 1$$

$$\lim _{h \rightarrow 0} \frac{1}{h}(f(x+h)-f(x))=\lim _{h \rightarrow 0} \frac{1}{h}\left(\frac{1}{x+h-1}-\frac{1}{x-1}\right)$$

$$=\lim _{h \rightarrow 0} \frac{1}{h}\left(\frac{x-1-(x+h-1)}{(x+h-1)(x-1)}\right)=\lim _{h \rightarrow 0} \frac{1}{h}\left(\frac{x-1-x-h+1}{(x+h-1)(x-1)}\right)$$

$$=\lim _{h \rightarrow 0} \frac{1}{h}\left(\frac{-h}{(x+h-1)(x-1)}\right)=\lim _{h \rightarrow 0} \frac{-1}{(x+h-1)(x-1)}=\frac{-1}{(x-1)(x-1)}=\frac{-1}{(x-1)^{2}}$$

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