Question

Given, f ^'(c)= lim x tends to c [{ f(x)- f(c)}/(x-c)] , when the limit exists . Usiing this definition investigate whether f(x)=1/x has a derivative at x=2 & obtain the derivative , if it exists?

Answer 1

$$\begin{gathered} \mathrm{Hi}, \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}} \\ \mathrm{at}\mathrm{x}=2\mathrm{here}\mathrm{c}=2 \\ \mathrm{f}\text{'}\left(2\right)=\underset{\mathrm{x}\to 2}{\lim }\frac{\frac{1}{\mathrm{x}}-{\displaystyle \frac{1}{2}}}{\mathrm{x}-2} \\ \mathrm{at}\mathrm{x}=2\mathrm{it}\mathrm{is}0/0\mathrm{form}\mathrm{so}\mathrm{using}\mathrm{L}\text{'}\mathrm{hospital}\mathrm{rule} \\ =\underset{\mathrm{x}\to 2}{\lim }\frac{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\frac{1}{\mathrm{x}}-{\displaystyle \frac{1}{2}}\right)}{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{x}-2\right)}=\underset{\mathrm{x}\to 2}{\lim }\frac{{\displaystyle \left(-\frac{1}{{\mathrm{x}}^2}\right)}}{{\displaystyle \left(1\right)}}=\frac{-1}{4} \\ \mathrm{so}\mathrm{limit}\mathrm{exists}\mathrm{so}\mathrm{its}\mathrm{derivative}\mathrm{is} \\ =\frac{-1}{{\mathrm{x}}^2} \end{gathered}$$