Question

If $f\left(1\right)=2andf\text{'}\left(1\right)=1$, then the value of $li{m}_{x\to 1}[2x-f(x\left)\right]/(x-1)$is

Answer 1

Finding the value of $\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{1}}\mathbf{}\mathbf{[}\mathbf{2}\mathit{x}\mathbf{-}\mathit{f}\mathbf{}\mathbf{(}\mathit{x}\mathbf{\left)}\mathbf{\right]}\mathbf{}\mathbf{/}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{}$:

Given that $f\left(1\right)=2andf\text{'}\left(1\right)=1$

$$\begin{gathered} li{m}_{x\to 1}\frac{[2x-f(x\left)\right]}{x-1} \\ Atx=1,\left(\frac{0}{0}form\right) \\ L\text{'}hospitalsrule \\ li{m}_{x\to 1}\frac{2-f\text{'}\left(x\right)}{1}=2-1 \\ =1 \end{gathered}$$

Hence, the value of $\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{x}\mathbf{\to }\mathbf{1}}\mathbf{}\mathbf{[}\mathbf{2}\mathit{x}\mathbf{-}\mathit{f}\mathbf{}\mathbf{(}\mathit{x}\mathbf{\left)}\mathbf{\right]}\mathbf{}\mathbf{/}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{}$is $\mathbf{1}$.