Locally Linear L’Hôpital

I’ll begin with a lemma. (I like to do a lemma now and then if for no other reason than having an excuse to explain what a lemma is – a simple theorem that is used in proving the main theorem.)

Lemma: If two lines intersect on the x-axis, then for any x the ratio of their y-coordinates is equal to the ratio of their slopes.

Proof: Two lines with slopes of m1 and m2 that intersect at (a, 0) on the x-axis have equations y1 = m1(xa) and y2 = m2(xa). Then

\displaystyle \frac{{{y}_{1}}}{{{y}_{2}}}=\frac{{{m}_{1}}\left( x-a \right)}{{{m}_{2}}\left( x-a \right)}=\frac{{{m}_{1}}}{{{m}_{2}}}

L’Hôpital’s Rule (Theorem): If f and g are differentiable near x = a and f(a) = g(a) = 0, then

\displaystyle \underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\frac{{f}'\left( x \right)}{{g}'\left( x \right)}

if the limit exists.

The fact that f(a) = g(a) = 0 means that the two functions intersect on the x-axis at (a, 0). For example, the functions f(x) = tan(x) and g(x) = sin(x) have this property at \left( \pi ,0 \right).

Figure 1. y = tan(x) in red and y = sin(x) in blue

Figure 1. y = tan(x) in red and y = sin(x) in blue

Now zoom-in several times centered at \left( \pi ,0 \right).

Figure 2. The previous graph zoomed in.

Figure 2. The previous graph zoomed in.

Whoa!

That looks like lines!!

It’s the local linearity property of differentiable functions – if you zoom-in enough any differentiable function eventually looks linear. So maybe near \left( \pi ,0 \right) the lemma applies. The only difference is that the slopes of the “lines” are the derivatives so

\displaystyle \underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\frac{{f}'\left( x \right)}{{g}'\left( x \right)}

 Now that does not quite prove L’Hôpital’s Rule, but it should give you and your students a good idea of why L’Hôpital’s Rule is true.

Then in our example:

\displaystyle \underset{x\to \pi }{\mathop{\lim }}\,\frac{\tan \left( x \right)}{\sin \left( x \right)}=\underset{x\to \pi }{\mathop{\lim }}\,\frac{\frac{1}{{{\left( \cos \left( x \right) \right)}^{2}}}}{\cos \left( x \right)}=\underset{x\to \pi }{\mathop{\lim }}\,\frac{1}{{{\left( \cos \left( x \right) \right)}^{3}}}=\frac{1}{{{\left( \cos \left( \pi \right) \right)}^{3}}}=-1

But isn’t that obvious from Figure 2?

Think about it.

More on this next time.

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