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5. Indeterminate Forms and L'Hôpital's Rule

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Some limit laws you may remember from calculus are the distributivity of limits over arithmetic operations: if \(\lim _{x\to a}f(x) = F\) and \(\lim _{x\to a}g(x) = G\), then one has \[\lim _{x\to a} \left( f(x) + g(x) \right) = F+G,\] and likewise when the \(+\) is replaced by a \(-\), \(\times\), or \(\div\). There is a caveat when doing division, and it’s that we need to avoid dividing by zero. For example, closing our eyes and applying the limit laws gives us \[\lim _{x\to 0} \frac{1}{x} = \frac{1}{0},\] and the limit does not exist. So this caveat isn’t much of a caveat after all.

There’s a caveat to this caveat, however, and it’s that division by zero doesn’t necessarily mean that a limit doesn’t exist. Consider, for instance, \[\lim _{x\to 0} \frac{x}{x} = 1.\] By blindly ignoring the warning not to divide by zero, we can apply the limit laws and say that \(\lim _{x\to 0} \frac{x}{x} = \frac{0}{0} = 1\). This looks reasonable enough: \(\frac{0}{0}=1\), since the zeroes cancel out.

Now consider doing the exact same thing with a slightly different function: \[\lim _{x\to 0 } \frac{x^2}{x} = \frac{0}{0} = 0.\] This looks reasonable as well: \(\frac{0}{0} = 0\) because zero divided by anything is \(0\). But this is incongruous with the previous value of \(\frac{0}{0}\) that we had arrived at above.

What’s happening is the expression \(\frac{0}{0}\) ignores features of the limits that affect its value. In \(\frac{x}{x}\), the numerator and denominator are “the same zero”, hence they can cancel out. However, in \(\frac{x^2}{x}\), the numerator is “more zero” than the denominator, overpowering it.

We say \(\frac{0}{0}\) is an indeterminate form, because it can have multiple different values depending on the context it is supressing. Likewise, the expression \(\frac{\infty}{\infty}\) is an indeterminate form — its value depends on “how infinite” the numerator and denominator are.

In these scenarios, we have been afforded the luxury of algebraic manipulations to simplify our limits and compute easily. However, there are more general scenarios where such luxuries do not exist, such as \[\lim _{x\to 0} \frac{e^x - 1}{\sin x}.\] Try as you might, no amount of algebra will save you here. This is what L’Hôpital’s rule is for: it’s a systematised way to compute limits that are indeterminate forms.

L’Hôpital’s rule is basically saying that “how zero” or “how infinite” a function is can be measured by its derivative. In the case that \(\lim _{x\to a}f(x) = f(a) = 0\), we have \(f(x) \approx f’(a) (x-a)\), and likewise \(g(x) \approx g’(a) (x-a)\). So we would expect \[\lim _{x\to a} \frac{f(x)}{g(x)} \approx \lim _{x\to a} \frac{f’(a)(x-a)}{g’(a)(x-a)} = \frac{f’(a)}{g’(a)}.\] Draw a picture to see what I mean! These are the linearisations of the two functions, and L’Hôpital’s rule simply makes this precise.

Returning to our example above, we can apply L’Hôpital’s Rule: \[ \lim _{x\to 0}\frac{e^x-1}{\sin x} = \lim _{x\to 0} \frac{e^x}{\cos x} = 1.\] Note L’Hôpital’s Rule applies here because our original limit looks like \(\frac{0}{0}\).

Other Indeterminate Forms: \(0\times\infty\)

Of course, \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\) aren’t the only indeterminate forms in the world. Another one is \(0\times \infty\), and an example of one we may wish to compute is \[\lim _{x\to\infty} x \ln x.\] I can’t apply L’Hôpital’s rule directly since it’s not \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), but I can turn it into this, either by doing the profoundly stupid argument \[\begin{align*} 0\times\infty = \frac{\infty}{\frac{1}{0}} = \frac{\infty}{\infty} && \textrm{or} && 0\times\infty = \frac{0}{\frac{1}{\infty}} = \frac{0}{0}.\end{align*}\] In this problem, this amounts to moving either the \(x\) or the \(\ln x\) to the denominator and flipping it, then applying L’Hôpital’s rule. \[\lim _{x\to 0}x\ln x = \lim _{x\to 0} \frac{\ln x}{\frac{1}{x}} = \lim _{x\to 0} \frac{\frac{1}{x}}{ - \frac{1}{x^2}} = \lim _{x\to 0} -x = 0. \]

Other Indeterminate Forms: \(0^0\) and \(1^\infty\)

As the section header suggests, the last two indeterminate forms to address are \(0^0\) and \(1^\infty\), and the example we’ll focus on is \[\lim _{x\to \infty} \left( 1+\frac{1}{x} \right)^x\] Taking the limit directly and pretending that the limit laws apply to exponentiation gives \(1^\infty\), and although it’s tempting to say that the answer is \(1\), one can check numerically that it’s not.

There was a government psyop conducted 2 weeks ago to inception the solution into your head. If we insert a natural log into the limit, it will pull apart the power and the base, and we can then use the other techniques we’ve learned about thus far! More specifically, if we give a name to the limit, say \[L = \lim _{x\to\infty} \left( 1+\frac{1}{x} \right)^x,\] we can take a natural log to get \[\ln L = \ln \left( \lim _{x\to\infty} \left( 1+\frac{1}{x} \right)^x \right) = \lim _{x\to\infty}\ln \left( \left( 1+\frac{1}{x} \right)^x \right) = \lim _{x\to\infty} x \ln \left( 1+\frac{1}{x} \right).\] Note we can swap \(\ln\) with \(\lim\) because \(\ln \) is continuous. Now this looks like \(\infty\times 0\), as \(\ln 1 = 0\), and we can do our usual tricks to apply L’Hôpital’s rule: \[\begin{align*} \ln L &= \lim _{x\to \infty} x \ln \left( 1+\frac{1}{x} \right) \\ &= \lim _{x\to\infty} \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}} \\ &= \lim _{x\to\infty} \frac{\frac{1}{1+\frac{1}{x}}\cdot \left( -\frac{1}{x^2} \right)}{-\frac{1}{x^2}} \\ &= \lim _{x\to\infty} \frac{1}{1+\frac{1}{x}} \\ \ln L &= 1.\end{align*}\] Don’t forget to undo the natural log at the end! Remember that we were looking for the limit \(L\), not its natural log. So, \[\lim _{x\to\infty} \left( 1+\frac{1}{x} \right)^x = L = e ^{\ln L} = e.\]