Limits and Towers

Consider the problem of finding the following limit:

\lim_{x \rightarrow 0} x^x

It’s actually not too bad. We can write

x^x = e^{ x \ln x}

and bring the limit into the exponent (as exponentiating is continuous) to get that

\lim_{x \rightarrow 0} x^x = e^{\lim_{x \rightarrow 0} x \ln x}

From here, all we need to do is calculate the remaining limit. There are a few ways to do this, but we opt for the powerful (and rather enjoyment-destroying!) l’Hôpital’s rule. We can write the limit as

\lim_{x \rightarrow 0} x \ln x = \lim_{x \rightarrow 0} \frac{\ln x}{1/x}.

We note that we are taking the limit of a fraction whose numerator and denominator are both tending to infinity. This means that we can use l’Hôpital’s rule, which allows us to differentiate both the numerator and denominator of the fraction and try the limit again. If we do this we get that

\lim_{x \rightarrow 0} x \ln x = \lim_{x \rightarrow 0} (-x) = 0.

Putting this back into the original problem, we have 

\lim_{x \rightarrow 0} x^x = e^0 = 1.

Ah, now the fun begins. Try your hand at finding the following limit.

\lim_{x \rightarrow 0} x^{x^x}

When you think you have the answer, you should look here to see if you’re correct and also for some super fun happy time.

 

One thought on “Limits and Towers

Leave a Reply to Mr. Anonymous. Cancel reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s