L'Hopital's Rule and Limits

We've used limits to find derivatives. Now we'll turn things upside-down and use derivatives to find limits, by way of a nive technique called L'Hopital's Rule.

Most of the limits we've looked at are naturally in one of the following forms:

• \(\lim_{x \to a}\frac{f(x)}{g(x)}\)
• \(\lim_{x \to a}(f(x) - g(x))\)
• \(\lim_{x \to a}f(x)g(x)\)
• \(\lim_{x \to a}f(x)^{g(x)}\)

Sometimes you can just substitute \(x = a\) and evaluate the limit directly, effectively using the continuity of \(f\) and \(g\). This method doesn't always work, for example:

\[ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \]

replacing \(x\) by 3 gives the indeterminate form \(\frac{0}{0}\).

It turns out that the first type, involving the ratio \(\frac{f(x)}{g(x)}\), is the most suitable for applying the rule, so we'll call it Type A. The next two types, involving \(f(x) - g(x)\) and \(f(x)g(x)\), both reduce directly to Type A, so we'll call them Type B1 and Type B2 respectively. Finally, we'll say that limits involving exponentials like \(f(x)^{g(x)}\) are Type C, since you can solve them by reducing them to Type B2 and then back to Type A.

Type A(\(0/0\) or \(\pm \infty/ \pm \infty\))

Consider limits of the form:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} \]

where \(f\) and \(g\) are nice differentiable functions. If \(g(a) \ne 0\), everything's great, you just substitude \(x = a\) to see the limit is \(\frac{f(a)}{g(a)}\).

The only other possibility is that \(f(a) = 0\) and \(g(a) = 0\). That is, the fraction \(\frac{f(a)}{g(a)}\) is the indeterminate form \(\frac{0}{0}\). Let's return to the definition of limits:

\[ f'(x) = lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

Since \(f\) and \(g\) are differentiable, we can find the linearization of both of them at \(x = a\). In fact, as we saw in the previous chapter, if \(x\) is close to \(a\), then:

\[ f(x) \cong f(a) + f'(a)(x - a) \]

and

\[ g(x) \cong g(a) + g'(a)(x - a) \]

Now we assume that \(f(a)\) and \(g(a)\) are both \(0\). This means:

\[ f(x) \cong f'(a)(x - a) \]

and

\[ g(x) \cong g'(a)(x - a) \]

If you divide the first equation by the second one, then assuming that \(x \ne a\), we get:

\[ \frac{f(x)}{g(x)} \cong \frac{f'(a)(x - a)}{g'(a)(x - a)} = \frac{f'(a)}{g'(a)} \]

The closer \(x\) is to \(a\), the better the approximation. This leads to one version of l'Hopital's Rule:

If \(f(a) = g(a) = 0\), then

\[ \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \]

Type B1(\(\infty - \infty\))

Here is a limit form:

\[ \lim_{x \to 0}(\frac{1}{sin(x)} - \frac{1}{x}) \]

As \(x \to 0^+\), both \(1/sin(x)\) and \(1/x\) go to \(\infty\). As \(x \to 0^-\), both quantities go to \(-\infty\). We can reduce this to Type A, just take a common denominator:

\[ \lim_{x \to 0}(\frac{1}{sin(x)} - \frac{1}{x}) = \lim_{x \to 0} \frac{x - sin(x)}{xsin(x)} \]

Now you can put \(x = 0\) and see \(0/0\) case. So we can apply l'Hopital's Rule:

\[ \lim_{x \to 0} (\frac{1}{sin(x)} - \frac{1}{x}) = \lim_{x \to 0} \frac{x - sin(x)}{xsin(x)} = \lim_{x \to 0}\frac{1 - cos(x)}{sin(x) + xcos(x)} = \lim_{x \to 0} \frac{sin(x)}{cos(x) + cos(x) - xsin(x)} = 0 \]

Type B2 (\(0 \times \pm \infty\))

Here's the example:

\[ \lim_{x \to 0^+} x\ln(x) \]

The idea is to move \(x\) into a new denominator by putting it there as \(\frac{1}{x}\):

\[ \lim_{x \to 0^+} x\ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{1/x} \]

Now the form is \(-\infty / \infty\), so we can use the l'Hopital's Rule:

\[ \lim_{x \to 0^+}x\ln(x) = \lim_{x \to 0^+}\frac{\ln(x)}{1/x} = \lim_{x \to 0^+}\frac{1/x}{-1/x^2} = 0 \]

Type C(\(1^{\pm \infty}, 0^0, \infty^0\))

Finally, the trickiest type involves limits like:

\[ \lim_{x \to 0^+} x^{sin(x)} \]

where both the base and exponent involve the dummy variable. The idea is to take the logarithm of the quatity \(x^{sin(x)}\) first, and work out its limit as \(x \to 0^+\):

\[ \lim_{x \to 0^+}ln(x^{sin(x)}) = \lim_{x \to 0^+}{sin(x)}\ln(x) = \lim_{x \to 0^+}ln(\frac{\ln(x)}{csc(x)}) = \lim_{x \to 0^+}ln(\frac{1/x}{-csc(x)cot(x)}) = ln(-1 \times 0) = ln(0) = 1 \]