I am asked to find the derivative function of [itex]f(x)=e^{x-2}[/itex] using the definition. That is to say, I have to evaluate this limit, if it exists: [tex]\lim_{x\rightarrow x_0}\frac{e^{x-2} - e^{x_0-2}}{x-x_0} = \lim_{x\rightarrow x_0}\frac{e^{x} - e^{x_0}}{e^2 (x-x_0)}[/tex] How can this undeterminate form be simplified? Thanks. (The answer is [itex]f'(x_0)=e^{x_0-2}[/itex].)
Factor out exp(x_{0}) from the numerator. Then expand exp(x-x_{0}) in a power series. The rest is obvious.
im used to defining derivatives in a slightly different way, but if you need to it should be easy to convert to your way: [tex]\frac{d}{dx}(e^{(x-2)}) = \lim_{h \rightarrow 0} \frac{e^{(x-2)+h} - e^{(x-2)}}{h}[/tex] [tex]= \lim_{h \rightarrow 0} e^{(x-2)} \frac{e^{h} -1}{h}[/tex] now noting that: [tex]e = \lim_{h \rightarrow 0} (1+h)^{1/h}[/tex] raising both sides to h (this is the only step I'm not sure about, but it gives the right answer): [tex]\lim_{h \rightarrow 0} e^h = \lim_{h \rightarrow 0} ((1+h)^{1/h})^h = \lim_{h \rightarrow 0} 1+h [/tex] so: [tex]\lim_{h \rightarrow 0} \frac{e^{h} -1}{h}= \lim_{h \rightarrow 0} \frac{1 + h -1}{h}= 1[/tex] which gives the answer.
lol, nobody is defining derivatives in different ways...only the notation differs [tex] h \equiv \Delta x \equiv x - x_0 [/tex]