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Show that if $h$ is near zero, $e^{h} \approx 1+h .$ Hint: Use (1 a) with $h$ replacing $u$.

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 3

The Number $e$

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University of Michigan - Ann Arbor

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this is the definition of E and since is a constant. We can take that limit and then we can take the power of age to both sides, which becomes a limit of age, power of age because the limit of one plus age. So when h takes very small numbers, this equation will hold approximately. This just gives us a result. I want to mention here that we can also derive this approximation from the Taylor expansions. If you know the tenor expansion, the expression of e to the power of X equals this infinite series and we can see that this approximation is just the first two terms of this series. One X is very small. X is very close to zero. This higher order decrease much faster than the first order. And this is why this approximation holds

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