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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