# Ask Uncle Colin: A Horrible-looking Limit

Dear Uncle Colin,

I need to work out $\lim_{x\to\infty} \left\{ \left[ \frac{ 2^{1/x} + 8^{1/x} }{2} \right]^x \right\}$. The textbook says it’s four, but my engineering friends all get different answers. How would you work it out?

- Giving Engineers Real Maths Analysis Is Naive

Hi, GERMAIN, and thanks for your message! This is one I Do Not Like The Look Of – but one I Stumbled On An Answer To.

### What does it remind me of?

Sometimes I think to try a thing, and realise that ‘why I thought to try that thing’ is probably not obvious, even to me. The thing I thought here was “can I make that into a hyperbolic function?”. The hint was the 2 on the bottom, which reminded me of sinh and cosh. The numbers on top are evenly spaced (geometrically!) around 4, as well - so it struck me as a thing to try.

(I also thought to expand the top binomially, before you start thinking I’m a genius or something. Not all of my ideas work.)

### Manipulating towards $\cosh(z)$

Just looking at the inner square bracket, $\left[ \frac{ 2^{1/x} + 8^{1/x} }{2} \right]^x$, I’m going to pull out $4^{1/x}$ and see what happens:

I get $\left[ 4^{1/x} \cdot \frac{ 2^{-1/x} + 2^{1/x}}{2} \right]^x$, which is $4 \left[ \frac{e^{-\ln(2) / x} + e^{\ln(x)/x}}{2}\right]^x$, or $4\cosh^x\left(\frac{\ln(2)}{x}\right)$.

Now the question is, what happens to $\cosh^x\left(\frac{\ln(2)}{x}\right)$ as $x$ goes to infinity. Ideally, it’ll be 1 - but how do we show that?

### In the limit

$\cosh(z)$ is an even function - which means its power series around $z=0$ consists only of even powers of $z$ - and it starts $1 + kz^2 + \dots$ for some constant of $k$ that I could work out, but won’t.

Here, I’m letting $z = \frac{\ln(2)}{x}$, which means that $\cosh(z)$ is $1 + k \frac{[\ln(2)]^2}{x^2} + \dots$, and that the first terms of $\cosh^x(z)$ are $1 + k\frac{[\ln(2)]^2}{x^2} + \dots$, which goes to 1 as $x$ goes to infinity.

We’ve still got the 4 we pulled out earlier, so our answer is $4\times 1 = 4$.

I suspect there are more rigorous ways to do the last part - but this is the way that made sense to me. I hope it helps!