# Ask Uncle Colin: A Fractional Limit

Dear Uncle Colin,

I need to find the limit as $x$ approaches 1 of $\frac{x^{29}-1}{x-1}$. I tried factoring out $x^{28}$ but didn't get anywhere.

- Learning How Others Proceed In This Awful Limit

Hi, LHOPITAL, and thanks for your message!

Factoring out an $x^{28}$ is very unlikely to get you anywhere, on the grounds that there isn't a factor of $x^{28}$! However, there are several excellent alternative approaches.

### Long division

The first, and perhaps most obvious, approach would be to divide the top by the bottom and see what shakes loose. That's quite a tedious thing, but after a while you spot that it's $x^{28} + x^{27} + x^{26} + ... + 1$. When $x=1$, each of those 29 terms evaluates to 1 and you're left with 29.

A similar approach spots that this is the sum of a geometric series with $a=1$, $r=x$ and $n=29$, and evaluates to the same thing.

### L'Hôpital's rule

An alternative that requires much less work is to use L'Hôpital's rule1 When you have $q(x)=\frac{f(x)}{g(x)}$ and want to find the limit of $q(x)$ at a value where $f(x)$ and $g(x)$ both evaluate to 0, it turns out the limit is $\frac{f'(x)}{g'(x)}$ evaluated at the same value of $x$.

Here, that gives $\frac{29x^{28}}{1}$, which is 29 when $x=1$.

### Binomial

A nice, if tricksy, method would be to let $x=y+1$.

This makes the expression $\frac{(y+1)^{28}-1}{y}$, and we're evaluating it where $y=0$.

The first term on the top starts $y^{28}+28y^{27}+...$ and ends $...+28y + 1$ - and that final one vanishes when you subtract the other 1 on the top of the fraction. That leaves the top with a common factor of $y$, cancelling neatly with the $y$ on the bottom.

That leaves $y^{27} + 28y^{26} + ... + 28$. Evaluated as $y$ approaches 02, this gives $y \to 28$.

Since $x=y+1$, that makes the limit we're looking for 29 again.

Hope that helps!

- Uncle Colin ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. discovered by a Bernoulli, of course. []
2. It fails at $y=0$ because we've divided by $y$, and we're not physicists, are we? []

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