Question

If $f\left(2\right)=4andf\text{'}\left(2\right)=4.$ Then, $li{m}_{x\to 2_{}}\frac{[xf(2)-2f(x\left)\right]}{(x-2)}=$

Answer 1

Finding the value of :$li{m}_{x\to 2_{}}\frac{[xf(2)-2f(x\left)\right]}{(x-2)}=$

Given that $f\left(2\right)=4andf\text{'}\left(2\right)=4.$ and

$li{m}_{x\to 2_{}}\frac{[xf(2)-2f(x\left)\right]}{(x-2)}$

If we apply limit value then the value will be indefinite form , $(\frac{0}{0}form)$

So ,Applying L 'hospital rule

$$\begin{gathered} li{m}_{x\to 2}\frac{[{\displaystyle \frac{d}{dx}}xf(2)-2{\displaystyle \frac{d}{dx}}f(x\left)\right]}{{\displaystyle \frac{d}{dx}}(x-2)} \\ =li{m}_{x\to 2}\frac{[4-2f\text{'}(x\left)\right]}{1}\mathbf{}\mathbf{(}\mathbf{}\mathbf{\because }\mathbf{}\mathbf{f}\mathbf{(}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{4}\mathbf{)} \\ =li{m}_{x\to 2}\frac{[4-2f\text{'}(x\left)\right]}{1} \\ =4-2f\text{'}\left(2\right)\mathbf{(}\mathbf{\because }\mathbf{}\mathbf{f}\mathbf{\text{'}}\mathbf{(}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{4}\mathbf{)} \\ =4-8 \\ =-4 \end{gathered}$$

Hence, the value of $li{m}_{x\to 2_{}}\frac{[xf(2)-2f(x\left)\right]}{(x-2)}$ is $\mathbf{-}\mathbf{4}$