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