Respuesta :

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

So, your limit is:

[tex] \lim_{h \to 0} \frac{x+h - x}{h} = \lim_{h \to 0} \frac{h}{h} [/tex]

Applying l'hopital's rule, 

[tex] \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} \frac{1}{1} = 1[/tex]

Giving a gradient of 1.

[tex]\begin{aligned}f'(x) &= \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} \\&= \lim_{h \to 0} \dfrac{(x+h) - x}{h} \\&=\lim_{h \to 0} \dfrac{h}{h} \\&=\lim_{h \to 0} (1) && (\text{\footnotesize since $h/h = 1$ for $h\ne 0$})\\&= 1\end{aligned}[/tex]

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.

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