- #1
autodidude
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I'm trying to differentiate e^x from first principles but I can't find a way to manipulate this expression [tex]\frac{e^h-1}{h}[/tex] so I can evaluate the limit without getting 0/0
Diffy said:
Diffy said:
eumyang said:Use the definition of e:
[itex]lim_{h \rightarrow 0} \left( 1 + h \right)^{1/h} = e[/itex]
So for small values of h,
[itex]e \approx \left( 1 + h \right)^{1/h}[/itex], or
[itex]e^h \approx 1 + h[/itex].
Replace eh in [itex]\frac{e^h-1}{h}[/itex] with 1 + h and go on from there.
Um, you need to simplify the expression.autodidude said:Don't we get 0/0 again?
Limit h->0
[tex]\frac{1+h-1}{h}[/tex]
The limit as h approaches 0 of (e^h-1)/h is an important concept in calculus and is used to find the instantaneous rate of change, or the slope of a curve, at a specific point on a function. It is also used in the definition of the derivative, which is a fundamental concept in calculus.
To evaluate this limit, you can use L'Hopital's rule, which states that if the limit of a function f(x) divided by g(x) as x approaches a is of the form 0/0 or infinity/infinity, then the limit can be evaluated by taking the derivative of both the numerator and denominator and then taking the limit again.
Yes, there are other methods for evaluating this limit, such as using Taylor series or the definition of the derivative. However, L'Hopital's rule is often the most efficient method.
The value of this limit is 1.
The value of this limit being 1 means that the slope of the tangent line at x=0 on the function f(x)=e^x is also 1. This is significant because it shows that the function e^x is its own derivative, and it has a constant rate of change of 1 at all points on the curve.