# How do you derive the maclaurin series for #(sin(4x^2))/x^3#?

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To derive the Maclaurin series for (\frac{\sin(4x^2)}{x^3}), we can use the Maclaurin series expansions for (\sin(x)) and (1/x^3).

- The Maclaurin series for (\sin(x)) is:

[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots ]

- The Maclaurin series for (1/x^3) is:

[ \frac{1}{x^3} = \frac{1}{x^3} ]

Now, we substitute these series expansions into the given expression:

[ \frac{\sin(4x^2)}{x^3} = \frac{\left(4x^2 - \frac{(4x^2)^3}{3!} + \frac{(4x^2)^5}{5!} - \frac{(4x^2)^7}{7!} + \cdots\right)}{x^3} ]

Simplify:

[ = 4 - \frac{64x^4}{3!} + \frac{1024x^{10}}{5!} - \frac{16384x^{16}}{7!} + \cdots ]

This is the Maclaurin series for (\frac{\sin(4x^2)}{x^3}).

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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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