Question

Sa se calculeze lim x-> 0 [f(x)-f(0)]/x unde este f:R-> R, f(x) = e^x - e^(-x)

Answer 1

Răspuns:

x→0 lim (f(x)-f(0)]/x =f `(x)

f `(x)=[eˣ-e⁻ˣ] `=eˣ-(-e⁻ˣ)=eˣ+e⁻ˣ

f `(0)=e⁰+e⁰=1+1=2

Explicație pas cu pas:

Answer 2

Răspuns:

2

Explicație pas cu pas:

Metoda 1:

Observam ca:

$\lim_{x \to 0} \frac{f(x)-f(0)}{x}= \lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=f'(0)$

Calculam f'(x):

$f'(x)=(e^x-e^{-x})'=(e^x)'-(e^{-x})'=e^x-e^{-x}*(-x)'=e^x-e^{-x}*(-1)=e^x+e^{-x}$

Acum putem calcula f'(0).

$f'(0)=e^0+e^0=1+1=2$

.

Metoda 2:

$\lim_{x \to 0} \frac{f(x)-f(0)}{x}= \lim_{x \to 0} \frac{e^x-e^{-x}-0}{x}= \lim_{x \to 0} \frac{e^x-e^{-x}}{x}=Cazul~\frac{0}{0}~si~aplicam~l'Hopital= \lim_{x \to 0} \frac{(e^x-e^{-x})'}{x'}= \lim_{x \to 0} \frac{e^x+e^{-x}}{1}= \frac{e^0+e^0}{1}=\frac{1+1}{1}=2$