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## Homework Statement

Use a series expansion to calculate L = [itex]\lim_{x\to\ 1}\frac{\sqrt[4]{80+x}-(3+\frac{(x-1)}{108})}{(x-1)^{2}}[/itex]

## Homework Equations

A function f(x)'s Taylor Series (if it exists) is equal to [itex]\sum_{n=0}^{\infty}\frac{f^{(n)}(x)}{n!}\cdot[/itex] (x-a)[itex]^{n}[/itex]

Newton's binomial theorem states that for all |x| < 1 and for any s we have [itex]\sum_{n=0}^{\infty}[/itex](s choose n)[itex]\cdot[/itex][itex]x^{n}[/itex]

## The Attempt at a Solution

This question is particularly aggravating since L'Hospital's rule applied twice consecutively yields L = [itex]\frac{-3}{32\cdot\sqrt[4]{81^{7}}}[/itex] = -[itex]\frac{1}{23328}[/itex], but that wouldn't be suitable given that no series expansion was used.

Using a Taylor expansion would require using an expansion point (a) different than 1, since it is that very point which we need in the first place. Using any other expansion point would require finding the series' radius of convergence. An alternative would be using L'Hospital's rule once, then trying to find that new limit's Taylor series. But don't you dare use L'Hospital's rule twice because that would give us the answer right away without using a series expansion

In any case, using a Taylor expansion sounds pretty desperate as the function's second, third, and fourth derivative are increasingly huge.

I couldn't get anywhere neither by rearranging the terms nor by using some form of substitution.

Help! :rofl:

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