Suppose $\underset{x\to 0}{\lim }\frac{g\left(x\right) - g\left(0\right)}{x}=1.$It follows necessarily that
1. g is not defined at x=0
2. g is not continuous at x=0
3. The limit of g(x) as x approaches 0 equals 1
4. g'(0)=1

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Please find below the solution to the asked query:

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{g}\left(\mathrm{x}\right)-\mathrm{g}\left(0\right)}{\mathrm{x}}=1 \\ \Rightarrow \underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{g}\left(\mathrm{x}\right)-\mathrm{g}\left(0\right)}{\mathrm{x}-0}=1...\left(\mathrm{i}\right) \\ \mathrm{By} \mathrm{definition} \mathrm{we} \mathrm{know} \mathrm{that} \\ \underset{\mathrm{x}\to \mathrm{a}}{\lim } \frac{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{a}\right)}{\mathrm{x}-\mathrm{a}}=\mathrm{f}\text{'}\left(\mathrm{a}\right) \\ \mathrm{Hence} \\ \underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{g}\left(\mathrm{x}\right)-\mathrm{g}\left(0\right)}{\mathrm{x}-0}=\mathrm{g}\text{'}\left(0\right)...\left(\mathrm{ii}\right) \\ \mathrm{By} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \mathrm{g}\text{'}\left(0\right)=1 \left[\mathrm{Answer}\right] \end{gathered}$$

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