# How can I show that #lim_(x->0)(1-cos(x^2))/(xsin(x^3))=1/2# using the Maclaurin series?

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Show that #lim_(x->0)(1-cos(x^2))/(xsin(x^3))=1/2# using the Maclaurin series. I know how maclaurin series works but that's ridiculous!

Show that

and:

so that:

Now consider the geometric series:

so that:

and then:

Finally:

and then:

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The terms cancel out on the numerator and denominator, so the denominator approaches 0. We don't need to consider the fact that the denominator is zero following direct substitution.

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To show that (\lim_{x \to 0} \frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{1}{2}) using the Maclaurin series, we can express (\cos(x)) and (\sin(x)) as their respective Maclaurin series expansions:

[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots ] [ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots ]

Substitute these series expansions into the given expression and simplify:

[ \frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{1 - (1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \ldots)}{x \cdot (x^3 - \frac{(x^3)^3}{3!} + \frac{(x^3)^5}{5!} - \frac{(x^3)^7}{7!} + \ldots)} ]

Now, expand the numerator and denominator:

[ \frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{\frac{x^4}{2!} - \frac{x^8}{4!} + \frac{x^{12}}{6!} - \ldots}{x^4 - \frac{x^{10}}{3!} + \frac{x^{16}}{5!} - \ldots} ]

Cancel out the common factors of (x^4):

[ \frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{\frac{1}{2!} - \frac{x^4}{4!} + \frac{x^8}{6!} - \ldots}{1 - \frac{x^6}{3!} + \frac{x^{12}}{5!} - \ldots} ]

Now, as (x) approaches 0, terms with higher powers of (x) become negligible, so the expression reduces to:

[ \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{2} ]

Therefore, (\lim_{x \to 0} \frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{1}{2}).

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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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