Simplifying the Limit for the Derivative of e^(x-2)

[quasar987]

I am asked to find the derivative function of f(x)=e^{x-2} using the definition. That is to say, I have to evaluate this limit, if it exists:

\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)}

How can this undeterminate form be simplified? Thanks.

(The answer is f'(x_0)=e^{x_0-2}.)

 

[mathman]

Factor out exp(x₀) from the numerator. Then expand exp(x-x₀) in a power series. The rest is obvious.

 

[quasar987]

Power serie not allowed, sorry.

 

[StatusX]

im used to defining derivatives in a slightly different way, but if you need to it should be easy to convert to your way:

\frac{d}{dx}(e^{(x-2)}) = \lim_{h \rightarrow 0} \frac{e^{(x-2)+h} - e^{(x-2)}}{h}

= \lim_{h \rightarrow 0} e^{(x-2)} \frac{e^{h} -1}{h}

now noting that:

e = \lim_{h \rightarrow 0} (1+h)^{1/h}

raising both sides to h (this is the only step I'm not sure about, but it gives the right answer):

\lim_{h \rightarrow 0} e^h = \lim_{h \rightarrow 0} ((1+h)^{1/h})^h = \lim_{h \rightarrow 0} 1+h

so:
\lim_{h \rightarrow 0} \frac{e^{h} -1}{h}= \lim_{h \rightarrow 0} \frac{1 + h -1}{h}= 1

which gives the answer.

 

[cepheid]

im used to defining derivatives in a slightly different way, but if you need to it should be easy to convert to your way:

lol, nobody is defining derivatives in different ways...only the notation differs

h \equiv \Delta x \equiv x - x_0

 

[StatusX]

yea, that's all i meant.

 

[quasar987]

This looks nice Status, but isn't there a e^{(x-2)} remaining?

 

[DeadWolfe]

Yes, which is multiplied by 1, giving the answer, as is easily verified using the chain rule.

 

[cepheid]

yea, that's all i meant.

oh ok, sorry