Evaluate a limit using a series expansion

[Perrault]

Homework Statement

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

Homework Equations

A function f(x)'s Taylor Series (if it exists) is equal to $\sum_{n=0}^{\infty}\frac{f^{(n)}(x)}{n!}\cdot$ (x-a)$^{n}$
Newton's binomial theorem states that for all |x| < 1 and for any s we have $\sum_{n=0}^{\infty}$(s choose n)$\cdot$$x^{n}$

The Attempt at a Solution

This question is particularly aggravating since L'Hospital's rule applied twice consecutively yields L = $\frac{-3}{32\cdot\sqrt[4]{81^{7}}}$ = -$\frac{1}{23328}$, 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!

 

[LCKurtz]

Homework Statement

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

Homework Equations

A function f(x)'s Taylor Series (if it exists) is equal to $\sum_{n=0}^{\infty}\frac{f^{(n)}(x)}{n!}\cdot$ (x-a)$^{n}$
Newton's binomial theorem states that for all |x| < 1 and for any s we have $\sum_{n=0}^{\infty}$(s choose n)$\cdot$$x^{n}$

The Attempt at a Solution

This question is particularly aggravating since L'Hospital's rule applied twice consecutively yields L = $\frac{-3}{32\cdot\sqrt[4]{81^{7}}}$ = -$\frac{1}{23328}$, 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!

Try writing $\sqrt[4]{80+x}$ as $(81 + (x-1))^\frac 1 4$ and expand that. You shouldn't need more than a few terms.

 

[Perrault]

Try writing $\sqrt[4]{80+x}$ as $(81 + (x-1))^\frac 1 4$ and expand that. You shouldn't need more than a few terms.

I'm sorry, I don't understand exactly what you are suggesting. How is $(81 + (x-1))^\frac 1 4$ any easier to expand than $\sqrt[4]{80+x}$?
Thanks for your help!

 

[SammyS]

I'm sorry, I don't understand exactly what you are suggesting. How is $(81 + (x-1))^\frac 1 4$ any easier to expand than $\sqrt[4]{80+x}$?
Thanks for your help!

Well, then x always appears as (x-1) . You could perhaps, substitute u = x-1 .

Look at the Taylor expansion of $\sqrt[4]{81+(x-1)}\,,$ or $\sqrt[4]{81+u}\ .$

Then rather than asking LCKurtz why he suggested the change, you can ask how he came up with such a useful suggestion.

 

[LCKurtz]

I'm sorry, I don't understand exactly what you are suggesting. How is $(81 + (x-1))^\frac 1 4$ any easier to expand than $\sqrt[4]{80+x}$?
Thanks for your help!

Just write out the first few terms of the binomial expansion for fractional exponents. You know, it starts with $81^{\frac 1 4}(x-1)^0 +\, ...$. Hopefully you know how to do that without using Taylor's expansion, although that will work.

 

[Perrault]

Thanks LCKurtz, it worked