How do you find the maclaurin series expansion of #1/sqrt(1x^2)#?
Construct the Maclaurin series for
#1/(sqrt(1x^2)) = sum_(k=0)^oo (prod_(i=1)^k (2k1))/(2^k k!) x^(2k)#
So:
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To find the Maclaurin series expansion of ( \frac{1}{\sqrt{1x^2}} ):

Start with the Maclaurin series expansion of ( (1 + t)^n ), which is ( \sum_{k=0}^{\infty} \binom{n}{k} t^k ), valid for ( t < 1 ).

Replace ( t ) with ( x^2 ) and ( n ) with ( \frac{1}{2} ) to obtain ( \frac{1}{\sqrt{1x^2}} = (1  x^2)^{\frac{1}{2}} ).

Apply the binomial series expansion to ( (1  x^2)^{\frac{1}{2}} ) to get the Maclaurin series expansion of ( \frac{1}{\sqrt{1x^2}} ).

Simplify the expression to obtain the final series expansion.

The resulting series will be the Maclaurin series expansion of ( \frac{1}{\sqrt{1x^2}} ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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