Question

Find, from the first principles, the gradient function of:

f(x)=x


`f'(x)= \lim_{h \to \ 0 (f(x+h)-f(x))/(h) }`

Answer 1

`\begin{aligned}f'(x) &= \lim_(h \to 0) (f(x+h) - f(x))/(h) \\&= \lim_(h \to 0) ((x+h) - x)/(h) \\&=\lim_(h \to 0) (h)/(h) \\&=\lim_(h \to 0) (1) && (\text{\footnotesize since $h/h = 1$ for $h\\e 0$})\\&= 1\end{aligned}`

h/h = 1 is valid for h ≠ 0. The limit does not care about h at 0; it only cares about the values around it.

Answer 2

f(x+h) = x+h, and f(x) = x

So, your limit is:


`\lim_(h \to 0) (x+h - x)/(h) = \lim_(h \to 0) (h)/(h)`

Applying l'hopital's rule,


`\lim_(h \to 0) (h)/(h) = \lim_(h \to 0) (1)/(1) = 1`

Giving a gradient of 1.