# How do you find the Maclaurin series of #f(x)=cosh(x)# ?

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To find the Maclaurin series of ( f(x) = \cosh(x) ), you can use the definition of the Maclaurin series:

[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n ]

For ( f(x) = \cosh(x) ), the ( n )-th derivative of ( \cosh(x) ) evaluated at ( x = 0 ) is:

[ f^{(n)}(0) = \begin{cases} \cosh(0) = 1 & \text{if } n \text{ is even} \ \sinh(0) = 0 & \text{if } n \text{ is odd} \end{cases} ]

So, the Maclaurin series for ( \cosh(x) ) is:

[ \cosh(x) = \sum_{n=0}^{\infty} \frac{1}{(2n)!} x^{2n} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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