# Set up Taylor expansion #\bb\color(red)\text(formula)# for #sqrt(x)# around #a=2#?

##
Check my work please?

#f'=(1/2)x^(-1/2)#

#f''=(1/2)(-1/2)x^(-3/2)#

#f'''=(1/2)(-1/2)(-3/2)x^(-5/2)#

#f^4=(1/2)(-1/2)(-3/2)(-5/2)x^(-7/2)#

#f^5=(1/2)(-1/2)(-3/2)(-5/2)(-7/2)x^(-9/2)#

#f'(2)=(1/2)*2^(-1/2)#

#f''(2)=(1/2)(-1/2)*2^(-3/2)#

#f''(2)'=(1/2)(-1/2)(-3/2)*2^(-5/2)#

#f^4(2)=(1/2)(-1/2)(-3/2)(-5/2)*2^(-7/2)#

#f^5(2)=(1/2)(-1/2)(-3/2)(-5/2)(-7/2)*2^(-9/2)#

Okay, I don't know where to go from here.

but I DO know that the exponents increase in this order...

#(-1/2)+(-1)^\beta# (not sure if #\beta=n# yet)

Check my work please?

Okay, I don't know where to go from here.

but I DO know that the exponents increase in this order...

Thus,

to observe that

The factorials are too simple for us to get any simpler.

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The Taylor expansion formula for ( \sqrt{x} ) around ( a = 2 ) is given by:

[ \sqrt{x} = \sqrt{a} + \frac{1}{2\sqrt{a}}(x - a) - \frac{1}{8a^{3/2}}(x - a)^2 + \frac{1}{16a^{5/2}}(x - a)^3 - \frac{5}{128a^{7/2}}(x - a)^4 + \cdots ]

This expansion can continue with higher-order terms if needed.

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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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