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.
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.